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

Answer: (d)

Answer: (d)

$x^2$ - 4x + 11

This can be written as

= $(x - 2)^2$ + 7

Here, $(x - 2)^2$ ≥ 0

So, given function can be not be less than 7.

Answer: (b)

I. $16x^2$ + 20x + 6 = 0

⇒$8x^2$ + 10x + 3 = 0⇒(4x + 3) + (2x + 1) = 0

∴ x = - $3/4$ or -$1/2$

II. $10y^2$ + 38y + 24 = 0⇒$5y^2$ + 19y + 12 = 0

∴ (y + 3) (5y + 4) = 0

∴ y = -3 or -$4/5$

Hence, x > y

Answer: (a)

Let α, β be two roots of the equation $x^2$ + px + q = 0

According to question

β = $α^2$

Sum of roots = α + $α^2$ = -p.......(1)

Product of roots = $α^3$ = q ......(2)

Dividing (1) by (2) we get

${α (α + 1)}/{α^3} = {-p}/q$

${α + 1}/{α^2} = -p/q$

Cubing on both sides, we get

$q^3 (α + 1)^3 = (-p)^3 (α^2)^3$

$q^3 (α^3 + 1 + 3 α^2 + 3 α) = -p^3 α^6 [α^3 = q α^6 = q^2]$

$q^3 [q + 1 + 3 (α^2 + α)] = -p^3 q^2$

q[q + 1 + 3(-p)] = -$p^3$

$q^2 + q - 3pq = - p^3$

⇒ $p^3 + q^2$ + q = 3pq

∴ Option (a) is correct.

Answer: (b)

Let y = $2x^2$ + 5x + 5

= 2 $[x^2 + 5/2 x + {25}/{16}] + 5 - {25}/8 = 2[x + 5/4]^2 + {15}/8$

For minimum value, $(x + 5/4)$ = 0

∴ $y_{min} = {15}/8$