NCERT Solutions for Class 10 Mathematics Chapter 4: Quadratic Equations (Complete Guide)
Quadratic equations form the cornerstone of algebraic modeling in secondary mathematics, directly bridging linear equations to higher-order polynomials and functional calculus. In the CBSE Class 10 curriculum, Chapter 4: Quadratic Equations equips students with the algebraic tools required to solve second-degree polynomial equations over the real numbers ($\mathbb{R}$). Mastering this chapter is essential for solving word problems related to area, projectile motion, and speed-distance-time relationships. The chapter fundamentally explores the algebraic manipulation of the standard quadratic form and the analytical evaluation of the discriminant to determine the nature of roots.
Plaintext
+---------------------------------------------------------------------------------------------------+
| QUADRATIC EQUATION ARCHITECTURE |
| |
| Standard Form: ax^2 + bx + c = 0 (where a, b, c are real numbers, a != 0) |
| |
| +-----------------------------+-----------------------------+---------------------------------+ |
| | Factorisation Method | Completing the Square | Quadratic Formula | |
| | Split the middle term 'bx' | Convert to (x + p)^2 = q | x = [-b ± sqrt(b^2 - 4ac)] / 2a | |
| | into two parts whose sum is | Add and subtract (b/2a)^2 | (Sridharacharya's Formula) | |
| | 'b' and product is 'ac'. | to form a perfect square. | Direct substitution method. | |
| +-----------------------------+-----------------------------+---------------------------------+ |
+---------------------------------------------------------------------------------------------------+
A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$. Any real number $\alpha$ is called a root of the quadratic equation if $a\alpha^2 + b\alpha + c = 0$. The roots of the quadratic equation are identical to the zeroes of the corresponding quadratic polynomial $p(x) = ax^2 + bx + c$.
The most crucial analytical concept introduced in this chapter is the Discriminant ($D$), defined as $D = b^2 – 4ac$. The discriminant directly dictates the nature (real or imaginary) and multiplicity (distinct or equal) of the roots without actually solving the equation:
- Two distinct real roots: if $b^2 – 4ac > 0$.
- Two equal real roots (coincident): if $b^2 – 4ac = 0$.
- No real roots (imaginary): if $b^2 – 4ac < 0$.
According to the official CBSE assessment guidelines, solving quadratic word problems requires a rigorous, step-by-step approach:
- Explicit variable declaration (e.g., “Let the first number be $x$”).
- Mathematical formulation of the equation based on the given conditions.
- Rearrangement into the standard form $ax^2 + bx + c = 0$.
- Clear algebraic resolution (via splitting the middle term or quadratic formula).
- Rejection of invalid physical values (e.g., negative speed or length) with a stated reason.
- Final concluding statement with appropriate units.
| Concept / Formula | Algebraic Expression | Key CBSE Constraints & Conditions | Analytical Interpretation |
| Standard Form | $ax^2 + bx + c = 0$ | $a, b, c \in \mathbb{R}$, $a \neq 0$ | Highest degree of variable $x$ must be exactly $2$. |
| Quadratic Formula | $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$ | Applied when factorisation is difficult. | Yields two roots explicitly: $\alpha$ (using $+$) and $\beta$ (using $-$). |
| Discriminant ($D$) | $D = b^2 – 4ac$ | $D \ge 0$ for real roots. | Evaluates what lies under the square root in the formula. |
| Equal Roots Condition | $b^2 – 4ac = 0$ | $x = -\frac{b}{2a}$ | The parabola touches the $x$-axis at exactly one point. |
| Distinct Roots Condition | $b^2 – 4ac > 0$ | Roots are real and unequal. | The parabola intersects the $x$-axis at two distinct points. |
Chapter-End Exercises – Exercise 4.1 (Page No. 73)
Question 1. Check whether the following are quadratic equations:
(i) $(x + 1)^2 = 2(x – 3)$
(ii) $x^2 – 2x = (-2)(3 – x)$
(iii) $(x – 2)(x + 1) = (x – 1)(x + 3)$
(iv) $(x – 3)(2x + 1) = x(x + 5)$
(v) $(2x – 1)(x – 3) = (x + 5)(x – 1)$
(vi) $x^2 + 3x + 1 = (x – 2)^2$
(vii) $(x + 2)^3 = 2x(x^2 – 1)$
(viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$
(i) $(x + 1)^2 = 2(x – 3)$
Answer:
Expand the left side using the identity $(a + b)^2 = a^2 + 2ab + b^2$ and the right side by distribution:
$$\begin{aligned} x^2 + 2(x)(1) + 1^2 &= 2x – 6 \\ x^2 + 2x + 1 &= 2x – 6 \end{aligned}$$
Bring all terms to the left side:
$$\begin{aligned} x^2 + 2x – 2x + 1 + 6 &= 0 \\ x^2 + 7 &= 0 \end{aligned}$$
This equation can be written as $1x^2 + 0x + 7 = 0$, which is of the standard form $ax^2 + bx + c = 0$ (where $a = 1 \neq 0$).
Therefore, it is a quadratic equation.
(ii) $x^2 – 2x = (-2)(3 – x)$
Answer:
Expand the right side by distribution:
$$\begin{aligned} x^2 – 2x &= -6 + 2x \end{aligned}$$
Bring all terms to the left side:
$$\begin{aligned} x^2 – 2x – 2x + 6 &= 0 \\ x^2 – 4x + 6 &= 0 \end{aligned}$$
This equation is of the standard form $ax^2 + bx + c = 0$ (where $a = 1, b = -4, c = 6$).
Therefore, it is a quadratic equation.
(iii) $(x – 2)(x + 1) = (x – 1)(x + 3)$
Answer:
Expand both sides using binomial multiplication:
$$\begin{aligned} x(x + 1) – 2(x + 1) &= x(x + 3) – 1(x + 3) \\ x^2 + x – 2x – 2 &= x^2 + 3x – x – 3 \\ x^2 – x – 2 &= x^2 + 2x – 3 \end{aligned}$$
Subtract $x^2$ from both sides and bring all terms to the left:
$$\begin{aligned} -x – 2 – 2x + 3 &= 0 \\ -3x + 1 &= 0 \end{aligned}$$
This equation is of the form $bx + c = 0$ (degree is $1$, $a = 0$).
Therefore, it is not a quadratic equation (it is a linear equation).
(iv) $(x – 3)(2x + 1) = x(x + 5)$
Answer:
Expand both sides:
$$\begin{aligned} x(2x + 1) – 3(2x + 1) &= x^2 + 5x \\ 2x^2 + x – 6x – 3 &= x^2 + 5x \\ 2x^2 – 5x – 3 &= x^2 + 5x \end{aligned}$$
Bring all terms to the left side:
$$\begin{aligned} 2x^2 – x^2 – 5x – 5x – 3 &= 0 \\ x^2 – 10x – 3 &= 0 \end{aligned}$$
This is of the standard form $ax^2 + bx + c = 0$ (where $a = 1 \neq 0$).
Therefore, it is a quadratic equation.
(v) $(2x – 1)(x – 3) = (x + 5)(x – 1)$
Answer:
Expand both sides:
$$\begin{aligned} 2x(x – 3) – 1(x – 3) &= x(x – 1) + 5(x – 1) \\ 2x^2 – 6x – x + 3 &= x^2 – x + 5x – 5 \\ 2x^2 – 7x + 3 &= x^2 + 4x – 5 \end{aligned}$$
Bring all terms to the left side:
$$\begin{aligned} 2x^2 – x^2 – 7x – 4x + 3 + 5 &= 0 \\ x^2 – 11x + 8 &= 0 \end{aligned}$$
This is of the standard form $ax^2 + bx + c = 0$.
Therefore, it is a quadratic equation.
(vi) $x^2 + 3x + 1 = (x – 2)^2$
Answer:
Expand the right side using the identity $(a – b)^2 = a^2 – 2ab + b^2$:
$$\begin{aligned} x^2 + 3x + 1 &= x^2 – 4x + 4 \end{aligned}$$
The $x^2$ term cancels out from both sides:
$$\begin{aligned} 3x + 4x + 1 – 4 &= 0 \\ 7x – 3 &= 0 \end{aligned}$$
This is a linear equation (degree $1$, $a = 0$).
Therefore, it is not a quadratic equation.
(vii) $(x + 2)^3 = 2x(x^2 – 1)$
Answer:
Expand the left side using $(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$ and the right side by distribution:
$$\begin{aligned} x^3 + 2^3 + 3(x^2)(2) + 3(x)(2^2) &= 2x^3 – 2x \\ x^3 + 8 + 6x^2 + 12x &= 2x^3 – 2x \end{aligned}$$
Bring all terms to the left side:
$$\begin{aligned} x^3 – 2x^3 + 6x^2 + 12x + 2x + 8 &= 0 \\ -x^3 + 6x^2 + 14x + 8 &= 0 \end{aligned}$$
This is a cubic equation (highest degree is $3$).
Therefore, it is not a quadratic equation.
(viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$
Answer:
Expand the right side using $(a – b)^3 = a^3 – b^3 – 3a^2b + 3ab^2$:
$$\begin{aligned} x^3 – 4x^2 – x + 1 &= x^3 – 2^3 – 3(x^2)(2) + 3(x)(2^2) \\ x^3 – 4x^2 – x + 1 &= x^3 – 8 – 6x^2 + 12x \end{aligned}$$
The $x^3$ term cancels out from both sides. Bring remaining terms to the left:
$$\begin{aligned} -4x^2 + 6x^2 – x – 12x + 1 + 8 &= 0 \\ 2x^2 – 13x + 9 &= 0 \end{aligned}$$
This is of the standard form $ax^2 + bx + c = 0$ (where $a = 2 \neq 0$).
Therefore, it is a quadratic equation.
Question 2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is $528\text{ m}^2$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is $306$. We need to find the integers.
(iii) Rohan’s mother is $26\text{ years}$ older than him. The product of their ages (in years) $3\text{ years}$ from now will be $360$. We would like to find Rohan’s present age.
(iv) A train travels a distance of $480\text{ km}$ at a uniform speed. If the speed had been $8\text{ km/h}$ less, then it would have taken $3\text{ hours}$ more to cover the same distance. We need to find the speed of the train.
(i) Rectangular Plot Area
Answer:
Step 1: Variable Declaration
Let the breadth of the rectangular plot be $x\text{ metres}$.
According to the given condition, the length is one more than twice its breadth.
Length $= (2x + 1)\text{ metres}$.
Step 2: Equation Formulation
The area of a rectangle is given by the formula $\text{Area} = \text{Length} \times \text{Breadth}$.
We are given that the Area $= 528\text{ m}^2$.
$$\begin{aligned} (2x + 1) \times x &= 528 \\ 2x^2 + x &= 528 \\ 2x^2 + x – 528 &= 0 \end{aligned}$$
Final Statement:
The situation is mathematically represented by the quadratic equation $2x^2 + x – 528 = 0$, where $x$ is the breadth of the plot in metres.
(ii) Product of Consecutive Positive Integers [BOARD EXAM FAVORITE / CBSE 2020]
Answer:
Step 1: Variable Declaration
Let the first positive integer be $x$.
Since the integers are consecutive, the next positive integer is $(x + 1)$.
Step 2: Equation Formulation
We are given that the product of these two integers is $306$.
$$\begin{aligned} x \times (x + 1) &= 306 \\ x^2 + x &= 306 \\ x^2 + x – 306 &= 0 \end{aligned}$$
Final Statement:
The situation is mathematically represented by the quadratic equation $x^2 + x – 306 = 0$, where $x$ is the smaller integer.
(iii) Rohan’s Age Problem [BOARD EXAM FAVORITE / CBSE 2019, 2022]
Answer:
Step 1: Variable Declaration
Let Rohan’s present age be $x\text{ years}$.
Since his mother is $26\text{ years}$ older than him, Rohan’s mother’s present age $= (x + 26)\text{ years}$.
Step 2: Formulating Ages After 3 Years
Age of Rohan $3\text{ years}$ from now $= (x + 3)\text{ years}$.
Age of his mother $3\text{ years}$ from now $= (x + 26) + 3 = (x + 29)\text{ years}$.
Step 3: Equation Formulation
The product of their ages $3\text{ years}$ from now is $360$.
$$\begin{aligned} (x + 3)(x + 29) &= 360 \\ x(x + 29) + 3(x + 29) &= 360 \\ x^2 + 29x + 3x + 87 &= 360 \\ x^2 + 32x + 87 – 360 &= 0 \\ x^2 + 32x – 273 &= 0 \end{aligned}$$
Final Statement:
The situation is mathematically represented by the quadratic equation $x^2 + 32x – 273 = 0$, where $x$ is Rohan’s present age in years.
(iv) Train Speed-Distance Problem [BOARD EXAM FAVORITE / CBSE 2020, 2023]
Answer:
Step 1: Variable Declaration
Let the original uniform speed of the train be $x\text{ km/h}$.
Total distance to travel $= 480\text{ km}$.
Step 2: Formulating Time Expressions
Using the relation $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$:
Time taken at original speed, $t_1 = \frac{480}{x}\text{ hours}$.
If the speed is reduced by $8\text{ km/h}$, the new speed is $(x – 8)\text{ km/h}$.
Time taken at reduced speed, $t_2 = \frac{480}{x – 8}\text{ hours}$.
Step 3: Equation Formulation
The problem states that the train takes $3\text{ hours}$ more at the slower speed. Therefore, the difference in time is $3\text{ hours}$.
$$\begin{aligned} t_2 – t_1 &= 3 \\ \frac{480}{x – 8} – \frac{480}{x} &= 3 \end{aligned}$$
Divide the entire equation by $3$ to simplify:
$$\frac{160}{x – 8} – \frac{160}{x} = 1$$
Multiply by the LCM $x(x – 8)$ to clear denominators:
$$\begin{aligned} 160x – 160(x – 8) &= x(x – 8) \\ 160x – 160x + 1280 &= x^2 – 8x \\ 1280 &= x^2 – 8x \\ x^2 – 8x – 1280 &= 0 \end{aligned}$$
Final Statement:
The situation is mathematically represented by the quadratic equation $x^2 – 8x – 1280 = 0$, where $x$ is the original speed of the train in km/h.
Chapter-End Exercises – Exercise 4.2 (Page No. 76)
Question 1. Find the roots of the following quadratic equations by factorisation:
(i) $x^2 – 3x – 10 = 0$
(ii) $2x^2 + x – 6 = 0$
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
(iv) $2x^2 – x + \frac{1}{8} = 0$
(v) $100x^2 – 20x + 1 = 0$
(i) $x^2 – 3x – 10 = 0$
Answer:
We need to split the middle term ($-3x$) into two terms whose sum is $-3$ and whose product is $1 \times (-10) = -10$.
These numbers are $-5$ and $+2$.
$$\begin{aligned} x^2 – 5x + 2x – 10 &= 0 \\ x(x – 5) + 2(x – 5) &= 0 \\ (x – 5)(x + 2) &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} x – 5 = 0 &\implies x = 5 \\ x + 2 = 0 &\implies x = -2 \end{aligned}$$
Therefore, the roots are $5$ and $-2$.
(ii) $2x^2 + x – 6 = 0$
Answer:
We need two numbers whose sum is $+1$ and product is $2 \times (-6) = -12$.
These numbers are $+4$ and $-3$.
$$\begin{aligned} 2x^2 + 4x – 3x – 6 &= 0 \\ 2x(x + 2) – 3(x + 2) &= 0 \\ (x + 2)(2x – 3) &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} x + 2 = 0 &\implies x = -2 \\ 2x – 3 = 0 &\implies 2x = 3 \implies x = \frac{3}{2} \end{aligned}$$
Therefore, the roots are $-2$ and $\frac{3}{2}$.
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ [BOARD EXAM FAVORITE / CBSE 2019, 2021]
Answer:
We need two numbers whose sum is $+7$ and product is $(\sqrt{2}) \times (5\sqrt{2}) = 5 \times 2 = 10$.
These numbers are $+5$ and $+2$.
$$\begin{aligned} \sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} &= 0 \\ \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) &= 0 \\ (x + \sqrt{2})(\sqrt{2}x + 5) &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} x + \sqrt{2} = 0 &\implies x = -\sqrt{2} \\ \sqrt{2}x + 5 = 0 &\implies \sqrt{2}x = -5 \implies x = -\frac{5}{\sqrt{2}} \end{aligned}$$
Therefore, the roots are $-\sqrt{2}$ and $-\frac{5}{\sqrt{2}}$.
(iv) $2x^2 – x + \frac{1}{8} = 0$
Answer:
First, multiply the entire equation by $8$ to eliminate the fraction:
$$\begin{aligned} 8\left(2x^2 – x + \frac{1}{8}\right) &= 8(0) \\ 16x^2 – 8x + 1 &= 0 \end{aligned}$$
We need two numbers whose sum is $-8$ and product is $16 \times 1 = 16$.
These numbers are $-4$ and $-4$.
$$\begin{aligned} 16x^2 – 4x – 4x + 1 &= 0 \\ 4x(4x – 1) – 1(4x – 1) &= 0 \\ (4x – 1)(4x – 1) &= 0 \\ (4x – 1)^2 &= 0 \end{aligned}$$
Setting the repeated factor to zero:
$$4x – 1 = 0 \implies 4x = 1 \implies x = \frac{1}{4}$$
Therefore, the roots are $\frac{1}{4}$ and $\frac{1}{4}$ (repeated real roots).
(v) $100x^2 – 20x + 1 = 0$
Answer:
We need two numbers whose sum is $-20$ and product is $100 \times 1 = 100$.
These numbers are $-10$ and $-10$.
$$\begin{aligned} 100x^2 – 10x – 10x + 1 &= 0 \\ 10x(10x – 1) – 1(10x – 1) &= 0 \\ (10x – 1)(10x – 1) &= 0 \\ (10x – 1)^2 &= 0 \end{aligned}$$
Setting the repeated factor to zero:
$$10x – 1 = 0 \implies 10x = 1 \implies x = \frac{1}{10}$$
Therefore, the roots are $\frac{1}{10}$ and $\frac{1}{10}$.
Question 2. Solve the problems given in Example 1.
Example 1 (i): John and Jivanti together have $45$ marbles. Both of them lost $5$ marbles each, and the product of the number of marbles they now have is $124$. We would like to find out how many marbles they had to start with.
Example 1 (ii): A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $55$ minus the number of toys produced in a day. On a particular day, the total cost of production was $₹750$. We would like to find out the number of toys produced on that day.
(i) Marbles Problem
Answer:
Let the number of marbles John had initially be $x$.
Since John and Jivanti together had $45$ marbles, the number of marbles Jivanti had initially $= (45 – x)$.
After losing $5$ marbles each:
- Marbles left with John $= x – 5$
- Marbles left with Jivanti $= (45 – x) – 5 = 40 – x$
The product of the number of marbles they now have is $124$:
$$\begin{aligned} (x – 5)(40 – x) &= 124 \\ 40x – x^2 – 200 + 5x &= 124 \\ -x^2 + 45x – 200 – 124 &= 0 \\ -x^2 + 45x – 324 &= 0 \end{aligned}$$
Multiply by $-1$ to standardize:
$$x^2 – 45x + 324 = 0$$
Factorisation: Split $-45x$ using two numbers whose sum is $-45$ and product is $324$. The numbers are $-36$ and $-9$.
$$\begin{aligned} x^2 – 36x – 9x + 324 &= 0 \\ x(x – 36) – 9(x – 36) &= 0 \\ (x – 36)(x – 9) &= 0 \end{aligned}$$
Setting each factor to zero yields $x = 36$ or $x = 9$.
- If John had $36$ marbles, Jivanti had $45 – 36 = 9$ marbles.
- If John had $9$ marbles, Jivanti had $45 – 9 = 36$ marbles.
Final Statement:
They started with $36$ and $9$ marbles respectively.
(ii) Cottage Industry Toys Problem
Answer:
Let the number of toys produced on that day be $x$.
The cost of production of each toy $= (55 – x)$ rupees.
Total cost of production is the number of toys multiplied by the cost per toy:
$$\begin{aligned} x(55 – x) &= 750 \\ 55x – x^2 &= 750 \\ -x^2 + 55x – 750 &= 0 \end{aligned}$$
Multiply by $-1$:
$$x^2 – 55x + 750 = 0$$
Factorisation: Split $-55x$ using two numbers whose sum is $-55$ and product is $750$. The numbers are $-30$ and $-25$.
$$\begin{aligned} x^2 – 30x – 25x + 750 &= 0 \\ x(x – 30) – 25(x – 30) &= 0 \\ (x – 30)(x – 25) &= 0 \end{aligned}$$
Setting each factor to zero yields $x = 30$ or $x = 25$.
Final Statement:
The number of toys produced on that day is either $25$ or $30$.
Question 3. Find two numbers whose sum is $27$ and product is $182$. [BOARD EXAM FAVORITE / CBSE 2018, 2021]
Answer:
Let the first number be $x$.
Since the sum of the two numbers is $27$, the second number is $(27 – x)$.
According to the given condition, their product is $182$:
$$\begin{aligned} x(27 – x) &= 182 \\ 27x – x^2 &= 182 \\ x^2 – 27x + 182 &= 0 \end{aligned}$$
Factorisation: Split $-27x$ into two numbers whose sum is $-27$ and product is $182$. The numbers are $-13$ and $-14$.
$$\begin{aligned} x^2 – 13x – 14x + 182 &= 0 \\ x(x – 13) – 14(x – 13) &= 0 \\ (x – 13)(x – 14) &= 0 \end{aligned}$$
Setting each factor to zero yields $x = 13$ or $x = 14$.
If the first number is $13$, the second is $27 – 13 = 14$.
If the first number is $14$, the second is $27 – 14 = 13$.
Final Statement:
The required two numbers are $13$ and $14$.
Question 4. Find two consecutive positive integers, sum of whose squares is $365$. [BOARD EXAM FAVORITE / CBSE 2020, 2022, 2023]
Answer:
Let the first positive integer be $x$.
The consecutive positive integer is $(x + 1)$.
According to the given condition, the sum of their squares is $365$:
$$\begin{aligned} x^2 + (x + 1)^2 &= 365 \\ x^2 + x^2 + 2x + 1 &= 365 \\ 2x^2 + 2x + 1 – 365 &= 0 \\ 2x^2 + 2x – 364 &= 0 \end{aligned}$$
Divide the entire equation by $2$ to simplify:
$$x^2 + x – 182 = 0$$
Factorisation: Split $+x$ into two numbers whose sum is $1$ and product is $-182$. The numbers are $+14$ and $-13$.
$$\begin{aligned} x^2 + 14x – 13x – 182 &= 0 \\ x(x + 14) – 13(x + 14) &= 0 \\ (x + 14)(x – 13) &= 0 \end{aligned}$$
Setting each factor to zero gives $x = -14$ or $x = 13$.
Since the question specifies positive integers, we reject $x = -14$.
Therefore, $x = 13$.
The second consecutive positive integer is $x + 1 = 13 + 1 = 14$.
Final Statement:
The required two consecutive positive integers are $13$ and $14$.
Question 5. The altitude of a right triangle is $7\text{ cm}$ less than its base. If the hypotenuse is $13\text{ cm}$, find the other two sides. [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
Let the base of the right triangle be $x\text{ cm}$.
According to the condition, the altitude (perpendicular) is $7\text{ cm}$ less than the base, so altitude $= (x – 7)\text{ cm}$.
The hypotenuse is given as $13\text{ cm}$.
Using Pythagoras Theorem ($\text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2$):
$$\begin{aligned} x^2 + (x – 7)^2 &= 13^2 \\ x^2 + x^2 – 14x + 49 &= 169 \\ 2x^2 – 14x + 49 – 169 &= 0 \\ 2x^2 – 14x – 120 &= 0 \end{aligned}$$
Divide the entire equation by $2$ to simplify:
$$x^2 – 7x – 60 = 0$$
Factorisation: Split $-7x$ using numbers with sum $-7$ and product $-60$. The numbers are $-12$ and $+5$.
$$\begin{aligned} x^2 – 12x + 5x – 60 &= 0 \\ x(x – 12) + 5(x – 12) &= 0 \\ (x – 12)(x + 5) &= 0 \end{aligned}$$
Setting to zero gives $x = 12$ or $x = -5$.
Since a geometric length cannot be negative, we reject $x = -5$.
Therefore, base $x = 12\text{ cm}$.
Altitude $= x – 7 = 12 – 7 = 5\text{ cm}$.
Final Statement:
The base of the triangle is $12\text{ cm}$ and its altitude is $5\text{ cm}$.
Question 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If the total cost of production on that day was $₹90$, find the number of articles produced and the cost of each article.
Answer:
Let the number of pottery articles produced on that day be $x$.
According to the condition, the cost of production of each article is $3$ more than twice the number produced.
Cost of each article $= (2x + 3)$ rupees.
The total cost of production is the number of articles multiplied by the cost per article:
$$\begin{aligned} x(2x + 3) &= 90 \\ 2x^2 + 3x &= 90 \\ 2x^2 + 3x – 90 &= 0 \end{aligned}$$
Factorisation: We need two numbers whose sum is $+3$ and product is $2 \times (-90) = -180$. The numbers are $+15$ and $-12$.
$$\begin{aligned} 2x^2 + 15x – 12x – 90 &= 0 \\ x(2x + 15) – 6(2x + 15) &= 0 \\ (2x + 15)(x – 6) &= 0 \end{aligned}$$
Setting to zero yields $x = -\frac{15}{2}$ or $x = 6$.
Since the number of articles produced must be a positive integer, we reject $x = -\frac{15}{2}$.
Therefore, the number of articles produced is $x = 6$.
The cost of each article $= 2(6) + 3 = 12 + 3 = 15$ rupees.
Final Statement:
The number of articles produced is $6$ and the cost of each article is $₹15$.
Chapter-End Exercises – Exercise 4.3 (Page No. 88)
Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) $2x^2 – 3x + 5 = 0$
(ii) $3x^2 – 4\sqrt{3}x + 4 = 0$
(iii) $2x^2 – 6x + 3 = 0$
(i) $2x^2 – 3x + 5 = 0$
Answer:
Compare with the standard form $ax^2 + bx + c = 0$:
$$a = 2, \quad b = -3, \quad c = 5$$
Calculate the Discriminant ($D$):
$$\begin{aligned} D &= b^2 – 4ac \\ D &= (-3)^2 – 4(2)(5) \\ D &= 9 – 40 \\ D &= -31 \end{aligned}$$
Since $D < 0$, the equation has no real roots.
(ii) $3x^2 – 4\sqrt{3}x + 4 = 0$ [BOARD EXAM FAVORITE / CBSE 2021]
Answer:
Compare with standard form:
$$a = 3, \quad b = -4\sqrt{3}, \quad c = 4$$
Calculate the Discriminant ($D$):
$$\begin{aligned} D &= b^2 – 4ac \\ D &= (-4\sqrt{3})^2 – 4(3)(4) \\ D &= 16(3) – 48 \\ D &= 48 – 48 = 0 \end{aligned}$$
Since $D = 0$, the equation has two equal real roots.
The roots are given by $x = -\frac{b}{2a}$:
$$x = -\frac{-4\sqrt{3}}{2(3)} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$$
Therefore, the equal real roots are $\frac{2}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}}$.
(iii) $2x^2 – 6x + 3 = 0$ [BOARD EXAM FAVORITE / CBSE 2020, 2022]
Answer:
Compare with standard form:
$$a = 2, \quad b = -6, \quad c = 3$$
Calculate the Discriminant ($D$):
$$\begin{aligned} D &= b^2 – 4ac \\ D &= (-6)^2 – 4(2)(3) \\ D &= 36 – 24 = 12 \end{aligned}$$
Since $D > 0$, the equation has two distinct real roots.
Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:
$$\begin{aligned} x &= \frac{-(-6) \pm \sqrt{12}}{2(2)} \\ x &= \frac{6 \pm 2\sqrt{3}}{4} \\ x &= \frac{2(3 \pm \sqrt{3})}{4} \\ x &= \frac{3 \pm \sqrt{3}}{2} \end{aligned}$$
Therefore, the distinct real roots are $\frac{3 + \sqrt{3}}{2}$ and $\frac{3 – \sqrt{3}}{2}$.
Question 2. Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots.
(i) $2x^2 + kx + 3 = 0$
(ii) $kx(x – 2) + 6 = 0$
(i) $2x^2 + kx + 3 = 0$ [BOARD EXAM FAVORITE / CBSE 2018, 2022]
Answer:
Comparing with $ax^2 + bx + c = 0$:
$a = 2$, $b = k$, $c = 3$.
For a quadratic equation to have two equal real roots, its discriminant must be zero ($D = 0$).
$$\begin{aligned} b^2 – 4ac &= 0 \\ k^2 – 4(2)(3) &= 0 \\ k^2 – 24 &= 0 \\ k^2 &= 24 \\ k &= \pm\sqrt{24} \\ k &= \pm2\sqrt{6} \end{aligned}$$
Therefore, the values of $k$ are $2\sqrt{6}$ and $-2\sqrt{6}$.
(ii) $kx(x – 2) + 6 = 0$ [BOARD EXAM FAVORITE / CBSE 2019, 2023]
Answer:
Expand to put the equation in standard form:
$$kx^2 – 2kx + 6 = 0$$
Comparing with $ax^2 + bx + c = 0$:
$a = k$, $b = -2k$, $c = 6$.
For equal real roots, $D = 0$:
$$\begin{aligned} b^2 – 4ac &= 0 \\ (-2k)^2 – 4(k)(6) &= 0 \\ 4k^2 – 24k &= 0 \\ 4k(k – 6) &= 0 \end{aligned}$$
Setting each factor to zero gives $4k = 0 \implies k = 0$ or $k – 6 = 0 \implies k = 6$.
However, if $k = 0$, the equation $0x^2 – 0x + 6 = 0$ implies $6 = 0$, which is absurd. Furthermore, a quadratic equation requires $a \neq 0$. Thus, $k = 0$ is rejected.
Therefore, the only valid value is $k = 6$.
Question 3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\text{ m}^2$? If so, find its length and breadth.
Answer:
Let the breadth of the rectangular mango grove be $x\text{ metres}$.
According to the condition, the length is twice the breadth, so length $= 2x\text{ metres}$.
The area of the rectangular grove is given as $800\text{ m}^2$.
$$\begin{aligned} \text{Length} \times \text{Breadth} &= \text{Area} \\ (2x)(x) &= 800 \\ 2x^2 &= 800 \\ x^2 &= 400 \end{aligned}$$
Write in standard form:
$$x^2 – 400 = 0$$
Calculate Discriminant $D = 0^2 – 4(1)(-400) = 1600 > 0$. Since $D$ is positive, real roots exist, so it is possible to design the grove.
Solving for $x$:
$$x = \pm\sqrt{400} = \pm 20$$
Since breadth cannot be negative, $x = 20$.
Breadth $= 20\text{ m}$.
Length $= 2x = 2(20) = 40\text{ m}$.
Final Statement:
Yes, it is possible. The length is $40\text{ m}$ and the breadth is $20\text{ m}$.
Question 4. Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is $20\text{ years}$. Four years ago, the product of their ages in years was $48$. [BOARD EXAM FAVORITE / CBSE 2020]
Answer:
Let the present age of the first friend be $x\text{ years}$.
Since the sum of their ages is $20$, the present age of the second friend is $(20 – x)\text{ years}$.
Ages four years ago:
Age of first friend $= (x – 4)\text{ years}$.
Age of second friend $= (20 – x) – 4 = (16 – x)\text{ years}$.
According to the condition, the product of their ages four years ago was $48$:
$$\begin{aligned} (x – 4)(16 – x) &= 48 \\ 16x – x^2 – 64 + 4x &= 48 \\ -x^2 + 20x – 64 – 48 &= 0 \\ -x^2 + 20x – 112 &= 0 \end{aligned}$$
Multiply by $-1$ to standardize:
$$x^2 – 20x + 112 = 0$$
Calculate the Discriminant ($D$) to check for real solutions:
$$\begin{aligned} a &= 1, \quad b = -20, \quad c = 112 \\ D &= b^2 – 4ac \\ D &= (-20)^2 – 4(1)(112) \\ D &= 400 – 448 \\ D &= -48 \end{aligned}$$
Since $D < 0$, the quadratic equation has no real roots. Therefore, no real values of $x$ satisfy the given conditions.
Final Statement:
No, the given situation is not possible.
Question 5. Is it possible to design a rectangular park of perimeter $80\text{ m}$ and area $400\text{ m}^2$? If so, find its length and breadth. [BOARD EXAM FAVORITE / CBSE 2023]
Answer:
Let the length of the park be $l$ and the breadth be $b$.
Given Perimeter $= 80\text{ m}$.
$$\begin{aligned} 2(l + b) &= 80 \\ l + b &= 40 \\ b &= 40 – l \end{aligned}$$
Given Area $= 400\text{ m}^2$.
$$\begin{aligned} l \times b &= 400 \\ l(40 – l) &= 400 \\ 40l – l^2 &= 400 \end{aligned}$$
Rearranging to standard quadratic form:
$$l^2 – 40l + 400 = 0$$
Calculate the Discriminant ($D$):
$$\begin{aligned} a &= 1, \quad b = -40, \quad c = 400 \\ D &= b^2 – 4ac \\ D &= (-40)^2 – 4(1)(400) \\ D &= 1600 – 1600 = 0 \end{aligned}$$
Since $D = 0$, real and equal roots exist. It is possible to design the park.
Solving for $l$:
$$l = \frac{-b}{2a} = \frac{-(-40)}{2(1)} = \frac{40}{2} = 20\text{ m}$$
Substitute $l = 20$ back to find breadth $b$:
$$b = 40 – 20 = 20\text{ m}$$
Since length equals breadth, the rectangular park is geometrically a square.
Final Statement:
Yes, it is possible. The length is $20\text{ m}$ and the breadth is $20\text{ m}$.
15 High-Yield Board Exam FAQs
FAQ 1 (HOTS / Upstream-Downstream). A motor boat whose speed is $18\text{ km/h}$ in still water takes $1\text{ hour}$ more to go $24\text{ km}$ upstream than to return downstream to the same spot. Find the speed of the stream. [CBSE 2019, 2022]
Answer:
Let the speed of the stream be $x\text{ km/h}$ ($x < 18$).
Speed of boat upstream $= (18 – x)\text{ km/h}$.
Speed of boat downstream $= (18 + x)\text{ km/h}$.
Time taken upstream $t_u = \frac{24}{18 – x}$.
Time taken downstream $t_d = \frac{24}{18 + x}$.
According to the problem, the difference in time is $1\text{ hour}$:
$$\begin{aligned} t_u – t_d &= 1 \\ \frac{24}{18 – x} – \frac{24}{18 + x} &= 1 \\ 24 \left[\frac{(18 + x) – (18 – x)}{(18 – x)(18 + x)}\right] &= 1 \\ 24 \left[\frac{2x}{324 – x^2}\right] &= 1 \\ 48x &= 324 – x^2 \\ x^2 + 48x – 324 &= 0 \end{aligned}$$
Factorisation: Split $48x$ into $54x$ and $-6x$.
$$\begin{aligned} x^2 + 54x – 6x – 324 &= 0 \\ x(x + 54) – 6(x + 54) &= 0 \\ (x + 54)(x – 6) &= 0 \end{aligned}$$
$x = -54$ (rejected since speed cannot be negative) or $x = 6$.
Therefore, the speed of the stream is $6\text{ km/h}$.
FAQ 2 (HOTS / Work-Rate Pipe Problem). Two water taps together can fill a tank in $9\frac{3}{8}\text{ hours}$. The tap of larger diameter takes $10\text{ hours}$ less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. [CBSE 2018, 2023]
Answer:
Total time taken together $= 9\frac{3}{8} = \frac{75}{8}\text{ hours}$. Total work done per hour $= \frac{8}{75}$.
Let the smaller tap take $x\text{ hours}$ to fill the tank.
Work done by smaller tap in $1\text{ hour} = \frac{1}{x}$.
The larger tap takes $(x – 10)\text{ hours}$.
Work done by larger tap in $1\text{ hour} = \frac{1}{x – 10}$.
$$\begin{aligned} \frac{1}{x} + \frac{1}{x – 10} &= \frac{8}{75} \\ \frac{x – 10 + x}{x(x – 10)} &= \frac{8}{75} \\ \frac{2x – 10}{x^2 – 10x} &= \frac{8}{75} \\ 75(2x – 10) &= 8(x^2 – 10x) \\ 150x – 750 &= 8x^2 – 80x \\ 8x^2 – 230x + 750 &= 0 \end{aligned}$$
Divide by $2$:
$$4x^2 – 115x + 375 = 0$$
Split $-115x$ into $-100x – 15x$:
$$\begin{aligned} 4x^2 – 100x – 15x + 375 &= 0 \\ 4x(x – 25) – 15(x – 25) &= 0 \\ (x – 25)(4x – 15) &= 0 \end{aligned}$$
$x = 25$ or $x = \frac{15}{4} = 3.75$.
If $x = 3.75$, the larger tap takes $3.75 – 10 = -6.25\text{ hours}$, which is impossible.
So $x = 25$.
Therefore, the smaller tap takes $25\text{ hours}$ and the larger tap takes $15\text{ hours}$.
FAQ 3 (Flight Delay Problem). A flight of $600\text{ km}$ was delayed due to bad weather. Its average speed for the trip was reduced by $200\text{ km/hr}$ and the time of flight increased by $30\text{ minutes}$. Find the original duration of the flight. [CBSE 2020, 2022]
Answer:
Let original uniform speed be $x\text{ km/h}$. Original time $t_1 = \frac{600}{x}\text{ hrs}$.
Reduced speed $= (x – 200)\text{ km/h}$. New time $t_2 = \frac{600}{x – 200}\text{ hrs}$.
Time difference $= 30\text{ minutes} = \frac{1}{2}\text{ hour}$.
$$\begin{aligned} t_2 – t_1 &= \frac{1}{2} \\ \frac{600}{x – 200} – \frac{600}{x} &= \frac{1}{2} \\ 600\left[\frac{x – (x – 200)}{x(x – 200)}\right] &= \frac{1}{2} \\ \frac{600 \times 200}{x^2 – 200x} &= \frac{1}{2} \\ x^2 – 200x &= 240000 \\ x^2 – 200x – 240000 &= 0 \end{aligned}$$
Split $-200x$ into $-600x + 400x$:
$$\begin{aligned} x(x – 600) + 400(x – 600) &= 0 \\ (x – 600)(x + 400) &= 0 \end{aligned}$$
$x = 600$ (reject $x = -400$). Original speed is $600\text{ km/h}$.
Original duration $t_1 = \frac{600}{600} = 1\text{ hour}$.
The original duration of the flight is $1\text{ hour}$.
FAQ 4 (HOTS / Number Reversal). A two-digit number is such that the product of its digits is $18$. When $63$ is subtracted from the number, the digits interchange their places. Find the number.
Answer:
Let the tens digit be $x$ and the units digit be $y$.
Original number $= 10x + y$.
Product of digits: $xy = 18 \implies y = \frac{18}{x}$.
Condition for reversal:
$$\begin{aligned} (10x + y) – 63 &= 10y + x \\ 9x – 9y – 63 &= 0 \\ x – y – 7 &= 0 \end{aligned}$$
Substitute $y = \frac{18}{x}$:
$$\begin{aligned} x – \frac{18}{x} – 7 &= 0 \\ x^2 – 18 – 7x &= 0 \\ x^2 – 7x – 18 &= 0 \end{aligned}$$
Factorisation:
$$\begin{aligned} x^2 – 9x + 2x – 18 &= 0 \\ x(x – 9) + 2(x – 9) &= 0 \\ (x – 9)(x + 2) &= 0 \end{aligned}$$
$x = 9$ (reject negative digit $-2$).
$y = \frac{18}{9} = 2$.
Therefore, the number is $92$.
FAQ 5 (Condition for Real Roots). Find the values of $k$ for which the quadratic equation $x^2 + 5kx + 16 = 0$ has no real roots.
Answer:
For no real roots, the discriminant must be strictly less than zero ($D < 0$).
$$a = 1, \quad b = 5k, \quad c = 16$$
$$\begin{aligned} b^2 – 4ac &< 0 \\ (5k)^2 – 4(1)(16) &< 0 \\ 25k^2 – 64 &< 0 \\ 25k^2 &< 64 \\ k^2 &< \frac{64}{25} \end{aligned}$$
Taking square roots yields the inequality:
$$-\frac{8}{5} < k < \frac{8}{5}$$
Therefore, the equation has no real roots when $k$ lies between $-\frac{8}{5}$ and $\frac{8}{5}$.
FAQ 6 (Assertion-Reasoning).
- Assertion (A): The equation $x^2 + x + 1 = 0$ has real roots.
- Reason (R): A quadratic equation $ax^2 + bx + c = 0$ has real roots if $b^2 – 4ac \ge 0$.
Answer:
Evaluate Discriminant for $x^2 + x + 1 = 0$:
$$D = (1)^2 – 4(1)(1) = 1 – 4 = -3$$
Since $D < 0$, the equation has no real roots. The Assertion is false.
The Reason correctly states the mathematical condition for real roots.
Therefore, Assertion (A) is false, but Reason (R) is true.
FAQ 7 (HOTS / Sum of Areas). Sum of the areas of two squares is $468\text{ m}^2$. If the difference of their perimeters is $24\text{ m}$, find the sides of the two squares. [CBSE 2019, 2021]
Answer:
Let the side of the larger square be $x$ and the smaller square be $y$.
Difference of perimeters:
$$\begin{aligned} 4x – 4y &= 24 \\ x – y &= 6 \implies x = y + 6 \end{aligned}$$
Sum of areas:
$$x^2 + y^2 = 468$$
Substitute $x$:
$$\begin{aligned} (y + 6)^2 + y^2 &= 468 \\ y^2 + 12y + 36 + y^2 &= 468 \\ 2y^2 + 12y – 432 &= 0 \\ y^2 + 6y – 216 &= 0 \end{aligned}$$
Factorisation: Split $6y$ into $18y – 12y$:
$$\begin{aligned} y(y + 18) – 12(y + 18) &= 0 \\ (y + 18)(y – 12) &= 0 \end{aligned}$$
$y = 12$ (reject $-18$).
$x = 12 + 6 = 18$.
Therefore, the sides are $18\text{ m}$ and $12\text{ m}$.
FAQ 8 (Geometry / Equilateral Triangle). If the area of an equilateral triangle is algebraically numerically equal to its perimeter, find the length of its side.
Answer:
Let the side of the equilateral triangle be $a$.
Area $= \frac{\sqrt{3}}{4}a^2$
Perimeter $= 3a$
Equating the two:
$$\begin{aligned} \frac{\sqrt{3}}{4}a^2 &= 3a \\ \frac{\sqrt{3}}{4}a^2 – 3a &= 0 \\ a\left(\frac{\sqrt{3}}{4}a – 3\right) &= 0 \end{aligned}$$
Since the side length cannot be zero, $a \neq 0$.
$$\begin{aligned} \frac{\sqrt{3}}{4}a – 3 &= 0 \\ \frac{\sqrt{3}}{4}a &= 3 \\ a &= \frac{12}{\sqrt{3}} = 4\sqrt{3} \end{aligned}$$
Therefore, the side of the equilateral triangle is $4\sqrt{3}\text{ units}$.
FAQ 9 (Solving by Quadratic Formula). Solve for $x$: $abx^2 + (b^2 – ac)x – bc = 0$. [CBSE 2020]
Answer:
We can use factorization by grouping.
$$abx^2 + b^2x – acx – bc = 0$$
Take common terms from the first two and last two:
$$\begin{aligned} bx(ax + b) – c(ax + b) &= 0 \\ (ax + b)(bx – c) &= 0 \end{aligned}$$
Setting each factor to zero:
$$\begin{aligned} ax + b &= 0 \implies x = -\frac{b}{a} \\ bx – c &= 0 \implies x = \frac{c}{b} \end{aligned}$$
Therefore, the roots are $-\frac{b}{a}$ and $\frac{c}{b}$.
FAQ 10 (HOTS / Fractional Equation). Solve for $x$: $\frac{1}{x + 4} – \frac{1}{x – 7} = \frac{11}{30}, x \neq -4, 7$. [CBSE 2018, 2023]
Answer:
Take LCM on the left side:
$$\begin{aligned} \frac{(x – 7) – (x + 4)}{(x + 4)(x – 7)} &= \frac{11}{30} \\ \frac{-11}{x^2 – 3x – 28} &= \frac{11}{30} \end{aligned}$$
Divide both sides by $11$:
$$\frac{-1}{x^2 – 3x – 28} = \frac{1}{30}$$
Cross multiply:
$$\begin{aligned} x^2 – 3x – 28 &= -30 \\ x^2 – 3x + 2 &= 0 \end{aligned}$$
Factorisation:
$$\begin{aligned} x^2 – 2x – x + 2 &= 0 \\ x(x – 2) – 1(x – 2) &= 0 \\ (x – 2)(x – 1) &= 0 \end{aligned}$$
Therefore, the roots are $x = 1$ and $x = 2$.
FAQ 11 (Age Problem with Reciprocals). The sum of the reciprocals of Rehman’s ages, (in years) $3\text{ years}$ ago and $5\text{ years}$ from now is $\frac{1}{3}$. Find his present age.
Answer:
Let present age be $x$.
Age $3\text{ years}$ ago $= (x – 3)$. Age $5\text{ years}$ hence $= (x + 5)$.
$$\begin{aligned} \frac{1}{x – 3} + \frac{1}{x + 5} &= \frac{1}{3} \\ \frac{(x + 5) + (x – 3)}{(x – 3)(x + 5)} &= \frac{1}{3} \\ \frac{2x + 2}{x^2 + 2x – 15} &= \frac{1}{3} \end{aligned}$$
Cross multiply:
$$\begin{aligned} 3(2x + 2) &= x^2 + 2x – 15 \\ 6x + 6 &= x^2 + 2x – 15 \\ x^2 – 4x – 21 &= 0 \end{aligned}$$
Factorisation:
$$(x – 7)(x + 3) = 0$$
Since age cannot be negative, $x = 7$.
Therefore, Rehman’s present age is $7\text{ years}$.
FAQ 12 (Parameter Finding with Distinct Roots). Find the values of $p$ for which the quadratic equation $px^2 – 2\sqrt{5}px + 15 = 0$ has two equal real roots. [CBSE 2019, 2022]
Answer:
Comparing with $ax^2 + bx + c = 0$, $a = p, b = -2\sqrt{5}p, c = 15$.
For equal real roots, $D = 0$:
$$\begin{aligned} b^2 – 4ac &= 0 \\ (-2\sqrt{5}p)^2 – 4(p)(15) &= 0 \\ 20p^2 – 60p &= 0 \\ 20p(p – 3) &= 0 \end{aligned}$$
$p = 0$ or $p = 3$.
Since $a \neq 0$ for a quadratic equation, $p = 0$ is rejected.
Therefore, the value of $p$ is $3$.
FAQ 13 (HOTS / Pythagoras Expansion). The hypotenuse of a right-angled triangle is $6\text{ m}$ more than twice the shortest side. If the third side is $2\text{ m}$ less than the hypotenuse, find the sides of the triangle.
Answer:
Let shortest side be $x$.
Hypotenuse $= 2x + 6$.
Third side $= (2x + 6) – 2 = 2x + 4$.
Using Pythagoras Theorem:
$$\begin{aligned} x^2 + (2x + 4)^2 &= (2x + 6)^2 \\ x^2 + 4x^2 + 16x + 16 &= 4x^2 + 24x + 36 \\ x^2 – 8x – 20 &= 0 \end{aligned}$$
Factorisation:
$$(x – 10)(x + 2) = 0$$
Since length must be positive, $x = 10$.
Shortest side $= 10\text{ m}$.
Third side $= 2(10) + 4 = 24\text{ m}$.
Hypotenuse $= 2(10) + 6 = 26\text{ m}$.
Therefore, the sides are $10\text{ m}$, $24\text{ m}$, and $26\text{ m}$.
FAQ 14 (Case Study: Quadratic Trajectory). A ball thrown vertically upward has its height $h$ (in meters) after $t$ seconds modeled by $h = -5t^2 + 20t$. Find the time taken for the ball to return to the ground.
Answer:
When the ball returns to the ground, its height $h = 0$.
$$\begin{aligned} -5t^2 + 20t &= 0 \\ -5t(t – 4) &= 0 \end{aligned}$$
This gives $t = 0$ (initial launch time) and $t = 4$.
Therefore, the ball returns to the ground after $4\text{ seconds}$.
FAQ 15 (Discriminant Analysis). State whether the quadratic equation $(x – \sqrt{2})^2 – 2(x + 1) = 0$ has real roots or not. Justify.
Answer:
Expand the equation:
$$\begin{aligned} (x^2 – 2\sqrt{2}x + 2) – 2x – 2 &= 0 \\ x^2 – x(2\sqrt{2} + 2) &= 0 \\ x[x – (2\sqrt{2} + 2)] &= 0 \end{aligned}$$
The roots are directly $x = 0$ and $x = 2\sqrt{2} + 2$, both of which are real numbers.
Alternatively, $D = (-(2\sqrt{2} + 2))^2 – 4(1)(0) = (2\sqrt{2} + 2)^2 > 0$, confirming real roots.
Therefore, yes, it has real roots.
To secure maximum marks in Class 10 Board examinations for Quadratic Equations, adhere strictly to the sequential algebraic methodology. Always define variables explicitly with their physical units before constructing equations from word problems. When writing standard equations, ensure terms are ordered from highest power to lowest ($ax^2 \to bx \to c$). Before executing the quadratic formula, always compute and explicitly state the Discriminant ($D = b^2 – 4ac$) separately; this often carries its own step mark and instantly reveals if a calculation error has yielded negative square roots in real-world contexts. Crucially, when evaluating real-world roots (speed, length, time, age), physically impossible negative or fractional values MUST be formally rejected with a written justification (e.g., “Since speed cannot be negative, we reject $x = -5$”). Ensure display equations are aligned clearly across multiple steps, mirroring the official marking scheme structure.