Quadratic Equations - Solved Examples

Exercise 4.2

Quadratic Equations Concepts

(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) $x^3 - 4x^2 - x + 1 = (x - 2)^3$

Solution

Standard form of a quadratic equation is $a{x^2} + bx + c = 0$ where $a ≠ 0$ and degree of quadratic equation is 2

(i) $(x + 1)^2=2(x - 3)$

$(x + 1)^2=2(x - 3)$
$⇒ x^2 + 2x + 1 = 2x - 6$
$⇒ x^2 + 2x + 1 - 2x + 6$
$⇒ x^2 + 7 = 0$

Here degree of $x^2 + 7 = 0$ is 2

∴ It is a quadratic equation.

(ii) $x^2 - 2x = (-2)(3 - x)$

$x^2 - 2x = (-2)(3 - x)$
$⇒ x^2 - 2x = -6 + 2x$
$⇒ x^2 - 2x + 6 - 2x = 0$
$⇒ x^2 - 4x + 6 = 0$

Here degree of $x^2 - 4x + 6 = 0$ is 2

∴ It is a quadratic equation.

(iii) $(x - 2)(x + 1) = (x - 1)(x + 3)$

$⇒ (x - 2)(x + 1) = (x - 1)(x + 3)$
$⇒ x^2 - 2x + x - 2 = x^2 - x + 3x - 3$
$⇒ x^2 - 2x + x - 2 - x^2 + x - 3x + 3 = 0$
$⇒ -3x + 1 = 0$

Here degree of $ -3x + 1 = 0$ is 1

∴ It is not a quadratic equation.

(iv) $(x - 3)(2x + 1) = x(x + 5)$

$⇒ (x - 3)(2x + 1) = x(x + 5)$
$⇒ 2x^2 - 6x + x - 3 = x^2 + 5x$
$⇒ 2x^2 - 6x + x - 3 - x^2 - 5x = 0$
$⇒ x^2 -10x - 3 = 0$

Here degree of $x^2 -10x - 3 = 0$ is 2

∴ It is a quadratic equation.

(v) $(2x - 1)(x - 3) = (x + 5)(x - 1)$

$⇒ (2x - 1)(x - 3) = (x + 5)(x - 1)$
$⇒ 2x^2 - 6x - x + 3 = x^2 + 5x - x - 5$
$⇒ 2x^2 - x^2 - 6x - x - 5x + x + 3 + 5 = 0 $
$⇒ x^2 - 11x + 8 = 0 $

Here degree of $x^2 - 11x + 8 = 0$ is 2

∴ It is a quadratic equation.

(vi) $x^2 + 3x + 1 = (x - 2)^2$

$⇒ x^2 + 3x + 1 = (x - 2)^2$
$⇒ x^2 + 3x + 1 = x^2 - 4x + 4$
$⇒ x^2 - x^2 + 3x + 4x + 1 - 4 = 0 $
$⇒ 7x - 3 = 0 $

Here degree of $7x - 3 = 0$ is 1

∴ It is not a quadratic equation.

(vii) $(x + 2)^3 = 2x(x^2 - 1)$

We know that (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

$(x + 2)^3 = 2x(x^2 - 1)$
$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$
$x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0$
$-x^3 + 6x^2 + 14x + 8 = 0$

Here degree of $-x^3 + 6x^2 + 14x + 8 = 0$ is 3

∴ It is not a quadratic equation.

(viii) $x^3 - 4x^2 - x + 1 = (x - 2)^3$

We know that (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

$x^3 - 4x^2 - x + 1 = (x - 2)^3$
$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$
$x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0$
$ 2x^2 - 13x + 9 = 0$

Here degree of $2x^2 - 13x + 9 = 0$ is 2

∴ It is a quadratic equation.

2

Represent the following situation in the form of quadratic equations.
(i) The area of a rectangular plot is $528m^2$. The length of the plot (in meters) 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 $480$ km at a uniform speed. If the speed had been $8$ 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.

Solution:

(i) Let breadth of the plot be $x$

It is given that the length of the plot (in meters) is one more than twice its breadth
∴ length of the plot will be $2x + 1$

Area of the rectangular plot = Length × Breadth = $528m^2$
∴ $x × (2x + 1) = 528m^2$
$2x^2 + x = 528$
$2x^2 + x - 528 = 0$

Hence, quadratic equation is $2x^2 + x - 528 = 0$

(ii) The product of two consecutive positive integers is $306$. We need to find the integers.

Let the first integer be $x$
Hence the next consecutive positive integer will be $x + 1$
Therefore,
$x(x + 1) = 306$
$x^2 + x - 306 = 0$

Hence, quadratic equation is $x^2 + x - 306 = 0$

(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.

Let Rohan's present age be $x$ years.
∴ Rohan's mother's age will be $x + 26$
Three years from now
Rohan's age will be $x + 3$ years and Rohan's mother's age will be $x + 26 + 3$

∴ $(x + 3)(x + 26 + 3) = 360$
∴ $(x + 3)(x + 29) = 360$
∴ $x^2 + 3x + 29x + 87 = 360$
∴ $x^2 + 32x + 87 - 360 = 0$
∴ $x^2 + 32x - 273 = 0$

Hence, quadratic equation is $x^2 + 32x - 273 = 0$

(iv) A train travels a distance of $480$ km at a uniform speed. If the speed had been $8$ 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.

Let the speed of train be $x$ km/h and the time it takes to travel be $t$
∴ Distance travelled = $ x × t $
⇒ $480 = x × t $ -----(i)

If the speed is reduced by 8 km/h, it takes 3 hours more to cover the same distance
∴ 480 = $ (x - 8) × (t + 3) $
⇒ $ (x - 8) × (t + 3) = 480$
⇒ $ xt - 8t + 3x - 24 = 480$

Using equation (i) to replace $t$ by $\dfrac{480}{x}$, we get,
⇒ $ 480 - 8 (\dfrac{480}{x}) + 3x - 24 = 480$
⇒ $ 3x^2 - 24x - 3840 = 0$

Dividing both sides by 3, we get, ⇒ $ x^2 - 8x - 1280 = 0$

Hence, quadratic equation is $x^2 - 8x - 1280 = 0$