quadratic equations Model Questions & Answers, Practice Test for ibps clerk prelims 2023

Answer: (b)

Equation, $ax^2$ + bx + c = 0 have root α and β then a $α^2$ + b α + c = 0 ⇒ a $α^2$ + b α = -c

⇒ (a α + b) = ${-c}/{α}$

$a β^2 + b β + c = 0 ⇒ a β^2 + b β = -c ⇒ (a β + b) = {- c}/{β}$

Now, $1/{(a α + b)} + 1/{(a β + b)} = - {α}/c - {β}/c$

= - ${(α + β)}/c = - {(- b/a)}/c = b/{ac}$

Answer: (d)

As α and β are roots of the equation $x^2 + kx - 15$ = 0

Then, sum of roots (α + β) = -k,

Product of roots α β = - 15 and (α - β) = 8 (given)

$(α - β)^2 + 4 α β = (α + β)^2$

64 + (4 × - 15) = $k^2$

⇒ $k^2$ = 4 ⇒ k = 2

Answer: (c)

Note: Let the quardatic equation be $ax^2$ + b + c = 0.

To find roots of this equation quickly, we find two factors of 'b' such that their sum is equal to b and their product is equal to the product of the coefficient of x2 and the constant term 'c'.

Let two such factors be α and β.

The α + β = b and α β = ca

In the second step, we divide these factors by the coefficient of $x^2$ ,

ie be 'a'.

In the next step, we change the signs of the outcome. These are the

roots of the equation.

Answer: (a)

Given, $1/{x + a} + 1/{x + b} = 1/c$

⇒ ${(x + b) + (x + a)}/{(x + a) (x + b)} = 1/c$

⇒ 2cx + (a + b)c = $x^2$ + (a + b)x + ab

⇒ $x^2$ + (a + b - 2c)x + ab - ac - bc = 0

Let the roots of above equation be α and β.

Given, α + β = 0

⇒ -a (a + b - 2c) = 0

⇒ a + b = 2c ....(i)

Now, α β = ab - ac - bc = ab - (a + b)c

= ab - (a + b) ${(a + b)}/2$ [from equation (i)]

= ${2ab - (a^2 + b^2 + 2ab)}/{2} = -{(a^2 + b^2)}/2$

Answer: (c)

Since, α and β are the roots of the equation

$x^2$ - 3x + 2 = 0

∴ α + β = 3 and α β = 2 ....(i)

Now, α + 1 + β + 1 = α + β + 2

= 3 + 2 = 5

and (α + 1) (β + 1) = α β + α + β + 1

2 + 3 + 1 = 6

Required equation is

$x^2$ - (α + 1 + β + 1) x + (α + 1) (β + 1) = 0

⇒ $x^2$ - 5x + 6 = 0