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

Answer: (d)

Given expression

$x^2$ - ax + b = 0

Here α and β are roots of the equation

∴ α + β = a and α β = b

S0, |α - β| = $√{(α + β)^2 - 4 α β} = √{a^2 - 4b}$

Answer: (d)

I. $5x^2$ – 18x + 9 = 0

⇒$5x^2$ – 15x – 3x + 9 = 0

⇒5x (x – 3) – 3 (x – 3) = 0

⇒(5x – 3) (x – 3)⇒x = $3/5$ or 3

Answer: (a)

We are given that $x^2 + 7x - 14 (k^2 - 7/8)$ = 0.

Let us check for the nature of roots.

We have,

D = $b^2 - 4ac = 7^2 - 4.1.14 (k^2 - 7/8)$

= 49 + $56k^2$ - 49 = 56

Now,

x = ${- b ± √D}/{2a} = {- 7 ± √{56k^2}}/2 = {-7 ± 2k √{14}}/2$

Since $√{14}$ is an irrational number, therefore any value of k will give an irrational number only.

Therefore, the given equation has no integral roots.

Answer: (a)

Let number of student in each row = x

and number of rows = y

Hence total number of students = xy

According to the question,

xy = (x + 1) (y - 2)

⇒ xy = xy - 2x + y - 2

⇒ 2x - y = -2

⇒ 2x - y = -2.......(i)

and xy = (x - 1) (y + 3)

⇒ xy = xy + 3x - y - 3

⇒ -3x + y = 3

3x - y = 3

From equation (i) and (ii)

x = 5, y = 12

Hence number students = xy

= 5 × 12 = 60

Answer: (b)