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

Answer: (b)

Roots of Aman (4, 3)

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

eq = $x^2$ - (α + β) x + α β

$x^2$ - 7x + 12

here constant 12 is wrong

Roots of Alok (3, 2)

equation = $x^2$ - (3 + 2)x + (3) (2)

= $x^2$ - 5x + 6

Here - 5 is wrong

So the correct equation is

$x^2$ - 7x + 6

Solution

$x^2$ - x - 6x - 16

x(x - 1) - 6(x - 1)

(x - 6) (x - 1)

x = 6 and x = 1

Answer: (a)

I. $8x^2$ + 6x – 5 = 0

⇒(4x + 5) (2x – 1) = 0 ∴ x = - $5/4$ or $1/2$

II. $12y^2$ – 22y + 8 = 0⇒$6y^2$ – 11y + 4 = 0

⇒(2y – 1) (3y – 4) = 0 ∴ y = $1/2$ or $4/3$

Hence, x ≤ y

Answer: (d)

According to question

Let one roots of equation is α then others roots of equation is $α^2$

∴ Sum of roots = α + $α^2 = -{(-b)}/1$

⇒ α (α + 1) = b......(i)

Product of roots = $α . α^2 = c/1$

⇒ $α^3 = c ⇒ α = c^{1/3}$

From equations (i) and (ii),

$c^{1/3} (c^{1/3} + 1)$ = b.....(iii)

On cubing both sides, we get

$c(c^{1/3} + 1)^3 = b^3$

⇒ $c[c + 1 + 3c^{1/3} (c^{1/3}) + 1] = b^3$

⇒ c(c + 1 + 3b) = $b^3$[from equation (iii)]

⇒ $b^3 = 3bc + c^2 + c$

Answer: (a)

Since the roots of the given equation are equal, therefore the discriminant of the given equation is zero. Thus,

$b^2 - 4ac = 0 ⇒ [b^2 (c - a)^2 - 4.a(b - c). c(a - b)] = 0$

⇒ $[b^2 (c^2 + a^2 - 2ac) - 4ac (ab - b^2 - ac + bc)]$ = 0

⇒ $[(ab)^2 + (bc)^2 + (-2ac)^2 + 2ab^2c - 4a^2bc - 4abc^2] = 0$

⇒ $[(ab)^2 + (bc)^2 + (-2ac)^2 + 2.ab.bc + 2.a.b. (-2ac) + 2.bc.(-2ac)] = 0$

⇒ $[b^2 c^2 + b^2 a^2 - 2ab^2c - 4a^2 bc + 4ab^2 c + 4a^2c^2 - 4abc^2] = 0$

⇒ $[b^2 a^2 + b^2 c^2 + 4a^2 c^2 - 2ab^2 c -4a^2 bc + 4ab^2c - 4abc^2]$ = 0

⇒ $(ab + bc - 2ac)^2$ = 0

⇒ ab + bc - 2ac = 0

⇒ ab + bc = 2ac

⇒ $b(a + c) = 2ac$

⇒ $2/b = {a + c}/{ac}$

⇒ $2/b = 1/c + 1/a$

Answer: (b)

The roots of equation

$(a^2 + b^2) x^2 - 2 (ac + bd)x + (c^2 + d^2)$ = 0 are equal.

$B^2 - 4AC = 0, B^2$ = 4AC

⇒ $4(ac + bd)^2 = 4(a^2 + b^2) (c^2 + d^2)$

⇒ $a^2 c^2 + b^2 d^2 + 2abcd = a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2$

⇒ $(ad - bc)^2$ = 0

∴ ad = bc