Solved Examples - Quadratic Equations - Quantitative Aptitude for Competitive

Section - 1
Distribute the following expressions.
Ques 1: (x+ 2)(x- 3)
Ans: (x + 2)(x - 3) = x(x - 3) + 2(x - 3)
= x² - 3x + 2x - 6
= x² - x - 6
Ques 2: (2s+1)(s + 5)
Ans: (2s + 1)(s + 5) = 2s(s + 5) + 1(s + 5)
= 2s² + 10s + s + 5
= 2s² + 11s + 5
Ques 3: (5 + a)(3 + a)
Ans: (5 + a)(3 + a) = 5·3 + 5a + 3a + a²
= 15 + 8a + a²
= a² + 8a + 15
Ques 4: (3 - z)(z + 4)
Ans: (3 - z)(z + 4) = 3(z + 4) - z(z + 4)
= 3z + 12 - z² - 4z
= -z² - z + 12
Section - 2
Ques 5: x² - 2x = 0
Ans: x² - 2x = 0
x(x - 2) = 0
Therefore x = 0 or x - 2 = 0 → x = 2
Ques 6: z² = -5 z
Ans: 

Bring all terms to one side:

z² + 5z = 0

Factor common term: z(z + 5) = 0

Therefore z = 0 or z + 5 = 0 → z = -5

Ques 7: y² + 4y + 3 = 0
Ans:

Factor the quadratic:

y² + 4y + 3 = (y + 1)(y + 3) = 0

Therefore y = -1 or y = -3

Ques 8: y²- 11y + 30 = 0
Ans:

Factor the quadratic:

y² - 11y + 30 = (y - 5)(y - 6) = 0

Therefore y = 5 or y = 6

Ques 9: y² + 3 y = 0
Ans:

Factor common term:

y(y + 3) = 0

Therefore y = 0 or y = -3

Ques 10: y² + 12y + 36 = 0
Ans:

Recognise a perfect square:

y² + 12y + 36 = (y + 6)² = 0

Therefore y + 6 = 0 → y = -6 (double root)

Section - 3
Simplify the following expressions.
Ques 11: 

The key to simplifying this expression is to recognise the special product in the numerator.

Replace the numerator by (a + b)(a - b) and then cancel the common factor (a - b) with the denominator:

(a + b)(a - b) ÷ (a - b) = a + b

After cancelling, the simplified result is a + b.

Ques 12: 

Ques 13: 

Ans: 

Because we have a common factor in both terms of the numerator, we can divide that factor out in order to simplify further. This is often a useful move when we are asked to add or subtract exponents with the same base:

Ques 14: 

Ans: It is tempting to expand the quadratic term in the numerator, but it is simpler to simplify first. Our goal is to cancel (2t - 1) from the denominator, so factor (2t - 1) from the numerator where possible.

One helpful method is to substitute x = 2t - 1 to see the structure more clearly:

Express the whole fraction in terms of x and simplify:

After cancellation and returning to t, we obtain:

x + 1 = (2t - 1) + 1 = 2t

Thus the simplified result is 2t.

Section - 4
Simplify the following expressions.
Ques 15: 

(A) 5 - x (B) x - 5 (C) x + 5
Ans: We can make this problem a lot simpler if we begin by factoring 3 out of both the numerator and the denominator:

The answers are not fractions, so we will have to remove the denominator by factoring the numerator as well:

Ques 16: 

Ans: This expression becomes manageable once we factor out the common term ab from the numerator. After factoring, the denominator reveals a familiar special product which can be cancelled:

Ques 17: 

Ans: There is no obvious single step to start with, so simplify before multiplying. Factor x³ from the numerator and x² from the denominator:

Cancel x² between numerator and denominator, leaving an x in the numerator. Factor x² - 1 in the numerator as (x - 1)(x + 1):

Multiply out the remaining factors in the numerator to compare with the answer choices: