1

Check if the following is a quadratic equation.

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

Solution $(2x - 1)(x - 3) = (x + 5)(x - 1)$
$⇒ (2x^2 - 6x - x + 3) = (x^2 - x + 5x - 5)$
$⇒ 2x^2 - 7x + 3 = x^2 + 4x - 5$
$⇒ 2x^2 - x^2 - 7x - 4x + 3 + 5 = 0$
$x^2 - 11x + 8 = 0$

This equation is of the form $ax^2 + bx + c = 0$, where a ≠ 0 and b, c are any real numbers.
Hence, this is a quadratic equation

2

Check if the following is a quadratic equation.

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

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

This equation is of the form $ax^2 + bx + c = 0$, where a ≠ 0 and b, c are any real numbers.
Hence, this is a quadratic equation

3

Simplify the equation $(2x - 1)^2 - 4x^2 + 5 = 0$ and show that it is not a quadratic equation.

Solution $(2x - 1)^2 - 4x^2 + 5 = 0$
$⇒ 4x^2 - 2x + 1 - 4x^2 + 5 = 0$
$⇒ - 2x + 6 = 0$

This equation is not of the form $ax^2 + bx + c = 0$, where a ≠ 0 and b, c are any real numbers.
Hence, this is a not quadratic equation

4

Raja and Raju together have 26 marbles. Both of them lost 3 marbles each and the product of the number of marbles they now have is 91. If we want to find out the number of marbles they had with them to start with, represent the above problem mathematically in terms of a quadratic equation.

Solution

Let the number of marbles with Raja by $x$ and the number of marbles with Raju be $y$.
The sum of the marbles they have is 26.
$⇒ x + y = 26$
After losing 3 marbles each, they will have $x - 3$ and $y - 3$ marbles respectively.
The product of the marbles they now have is 91
$⇒ (x - 3)(y - 3) = 91$
$⇒ xy - 3x - 3y + 9 = 91$
$⇒ xy - 3(x + y) + 9 = 91$
$⇒ xy - 3(x + y) = 91 - 9$
Replacing $(x + y)$ by 26, we get
$xy - 3 × 26 = 82$
$⇒ xy - 78 = 82$ or $xy = 160$
Replacing x = 26 - y
$(26 - y)y = 160$
$-y^2 + 26y - 160 = 0$ or $y^2 - 26y + 160 = 0$

5

Represent the following situation in the form of a quadratic equation
The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, then we need to find the sides of the field.

Solution

Let the shorter side be $x$.
∴ The diagonal is $(x + 60)$ and longer side is $(x + 30)$
Using Pythagoras Theorem,
$(x + 60)^2$ = $(x + 30)^2 + x^2$
$⇒ x^2 + 120x + 3600$ = $x^2 + 60x + 900 + x^2$
$⇒ x^2 - 2x^2 + 120x - 60x$ $+ 3600 - 900 = 0$
$⇒ -x^2 + 60x + 2700 = 0$
$⇒ x^2 - 60x - 2700 = 0$

6

Solve the following quadratic equation
$8x^2 - 22x - 21 = 0$

Solution

Given equation is $8x^2 - 22x - 21 = 0$
$⇒ 8x^2 - 28x + 6x - 21 = 0$
$[ ∵ -28 + 6 = -22$ and $-28 × 6 = -168]$
$⇒ 4x(2x - 7) + 3(2x - 7) = 0$
$⇒ (4x + 3)(2x - 7) = 0$
$∴ 4x + 3 = 0 ⇒ x = \dfrac{-3}{4}$ and $2x - 7 = 0 ⇒ x = \dfrac{7}{2}$
Hence, $x = \dfrac{-3}{4}$ and $x = \dfrac{7}{2}$ are the roots of the given equation.

7

The product of two consecutive integers is 20. Using quadratic equation, find the integers.

Solution

Let the two consecutive numbers be $x$ and $x+1$.
$(x) × (x+1) = 20$
$⇒ x^2 + x - 20 = 0$
$⇒ x^2 + 5x - 4x - 20 = 0$
$[ ∵ 5 - 4 = 1$ and $5 × -4 = -20]$
$⇒ x(x + 5) - 4(x + 5) = 0$
$⇒ (x + 5)(x - 4) = 0$
Hence $x = -5$ or $x = 4$
Hence 1st number is $-5$ and second number is $-4$ or 1st number is $4$ and 2nd number is $5$

8

Two numbers differ by 3 and their product is 504. Find the numbers.

Solution

Let the two numbers be $x$ and $x+3$.
$(x) × (x+3) = 504$
$⇒ x^2 + 3x - 504 = 0$
$⇒ x^2 + 24x - 21x - 504 = 0$
$[ ∵ 24 - 21 = 3$ and $24 × -21 = -504]$
$⇒ x(x + 24) - 21(x + 24) = 0$
$⇒ (x + 24)(x - 21) = 0$
Hence $x = -24$ or $x = 21$
Hence 1st number is $-24$ and second number is $-21$ or 1st number is $21$ and 2nd number is $24$

9

A train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/h from its usual speed. Find the usual speed of the train.

Solution

Let the average speed of the passenger train be $x$ km/h
Hence the time taken by the train to travel 300 km (t₁) = $\dfrac{300}{x}$
After increasing the speed by 5 km/h, the new speed will be $(x + 5)$ km/h
Time taken to cover 300 km with new speed (t₂) = $\dfrac{300}{x + 5}$ km/h
As t₁ - t₂ = 2

$⇒ \dfrac{300}{x} - \dfrac{300}{x + 5} = 2$
$⇒ \dfrac{(300)(x + 5) - 300x}{x^2 + 5x} = 2$
$⇒ 300x + 1500 - 300x = 2(x^2 + 5x)$
$⇒ 1500 = 2x^2 + 10x$
$⇒ 2x^2 + 10x - 1500 = 0$
$⇒ 2x^2 + 60x - 50x - 1500 = 0$
$⇒ 2x(x + 30) - 50(x + 30) = 0$
$⇒ (x + 30)(2x - 50) = 0$
Hence $x = -30$ or $x = 25$

As speed cannot be negative, the usual spped of the train is 25 km/h

10

Find the values of $k$ for which the given equation has real and equal roots

$x^2 + k(4x + k - 1) + 2 = 0$

Solution

$x^2 + k(4x + k - 1) + 2 = 0$
$⇒ x^2 + 4kx + k^2 - k + 2 = 0$
Comparing this equation with $ax^2 + bx + c = 0$,
$a = 1, b = 4k$ and $c= k^2 - k + 2$
$⇒$ Discriminant $D = b^2 - 4ac$
$16k^2 - 4 × 1 × (k^2 - k + 2)$
$16k^2 - 4k^2 + 4k - 8$
$12k^2 + 4k -8$
For equal roots, D = 0 ⇒ $12k^2 + 4k -8 = 0$
$⇒ 12k^2 + 12k - 8k - 8 = 0$
$⇒ 12k(k + 1) - 8(k + 1) = 0$
$⇒ (k + 1)(12k - 8) = 0$
Hence, for equal roots, $k = -1$ or $k = \dfrac{2}{3}$