NCERT Solutions Class 10 Maths Chapter 4: Quadratic Equations
Detailed solutions for Exercise 4.1 (Introduction to Quadratic Equations).
#### **Q1. 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)** $x^3 – 4x^2 – x + 1 = (x – 2)^3$
#### **A1. Solution:**
**(i) Solution:**
$(x + 1)^2 = 2(x – 3)$
$x^2 + 2x + 1 = 2x - 6$
$x^2 + 2x - 2x + 1 + 6 = 0$
$x^2 + 7 = 0$
Since it is of the form $ax^2 + bx + c = 0$ (where $a \neq 0$), **it is a quadratic equation.**
**(ii) Solution:**
$x^2 – 2x = (–2) (3 – x)$
$x^2 - 2x = -6 + 2x$
$x^2 - 2x - 2x + 6 = 0$
$x^2 - 4x + 6 = 0$
**It is a quadratic equation.**
**(iii) Solution:**
$(x – 2)(x + 1) = (x – 1)(x + 3)$
$x^2 + x - 2x - 2 = x^2 + 3x - x - 3$
$x^2 - x - 2 = x^2 + 2x - 3$
Cancel $x^2$ from both sides:
$-x - 2x - 2 + 3 = 0$
$-3x + 1 = 0$
This is a linear equation, not quadratic.
**It is NOT a quadratic equation.**
**(iv) Solution:**
$(x – 3)(2x + 1) = x(x + 5)$
$2x^2 + x - 6x - 3 = x^2 + 5x$
$2x^2 - 5x - 3 - x^2 - 5x = 0$
$x^2 - 10x - 3 = 0$
**It is a quadratic equation.**
**(v) Solution:**
$(2x – 1)(x – 3) = (x + 5)(x – 1)$
$2x^2 - 6x - x + 3 = x^2 - x + 5x - 5$
$2x^2 - 7x + 3 = x^2 + 4x - 5$
$2x^2 - x^2 - 7x - 4x + 3 + 5 = 0$
$x^2 - 11x + 8 = 0$
**It is a quadratic equation.**
**(vi) Solution:**
$x^2 + 3x + 1 = (x – 2)^2$
$x^2 + 3x + 1 = x^2 - 4x + 4$
Cancel $x^2$:
$3x + 4x + 1 - 4 = 0$
$7x - 3 = 0$
**It is NOT a quadratic equation.**
**(vii) Solution:**
$(x + 2)^3 = 2x (x^2 – 1)$
$x^3 + 8 + 6x(x+2) = 2x^3 - 2x$ (Using $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$)
$x^3 + 8 + 6x^2 + 12x = 2x^3 - 2x$
$-x^3 + 6x^2 + 14x + 8 = 0$
Degree is 3 (Cubic equation).
**It is NOT a quadratic equation.**
**(viii) Solution:**
$x^3 – 4x^2 – x + 1 = (x – 2)^3$
$x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x(x-2)$
$x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x^2 + 12x$
Cancel $x^3$:
$-4x^2 + 6x^2 - x - 12x + 1 + 8 = 0$
$2x^2 - 13x + 9 = 0$
**It is a quadratic equation.**
#### **Q2. 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 $26$ years older than him. The product of their ages (in years) $3$ 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 $3$ hours more to cover the same distance. We need to find the speed of the train.
#### **A2. Solution:**
**(i) Solution:**
Let the breadth of the plot be $x$ meters.
The length is one more than twice the breadth: $2x + 1$.
Area = Length $\times$ Breadth = $528$
$(2x + 1)x = 528$
$2x^2 + x = 528$
**Quadratic Equation:** $2x^2 + x - 528 = 0$
**(ii) Solution:**
Let the two consecutive integers be $x$ and $x + 1$.
Product = $306$
$x(x + 1) = 306$
$x^2 + x = 306$
**Quadratic Equation:** $x^2 + x - 306 = 0$
**(iii) Solution:**
Let Rohan's present age be $x$ years.
Mother's age = $x + 26$ years.
**3 years from now:**
Rohan's age = $x + 3$
Mother's age = $(x + 26) + 3 = x + 29$
Product = $360$
$(x + 3)(x + 29) = 360$
$x^2 + 29x + 3x + 87 = 360$
$x^2 + 32x + 87 - 360 = 0$
**Quadratic Equation:** $x^2 + 32x - 273 = 0$
**(iv) Solution:**
Let the speed of the train be $x$ km/h.
Distance = $480$ km.
Time taken at original speed = $\frac{480}{x}$ hours.
New speed = $(x - 8)$ km/h.
Time taken at new speed = $\frac{480}{x - 8}$ hours.
Difference in time = $3$ hours.
$\frac{480}{x - 8} - \frac{480}{x} = 3$
$\frac{480x - 480(x - 8)}{x(x - 8)} = 3$
$\frac{480x - 480x + 3840}{x^2 - 8x} = 3$
$\frac{3840}{x^2 - 8x} = 3$
Divide both sides by 3:
$\frac{1280}{x^2 - 8x} = 1$
$x^2 - 8x = 1280$
**Quadratic Equation:** $x^2 - 8x - 1280 = 0$