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

Answer: (d)

Let α and β be the roots of this quadratic equation

$2x^2$ - 11x + 5 = 0

∴ α + β = - ${(- 11)}/2 = {11}/2$.......(i)

and α . β = $5/2$

We know that

$(α - β)^2 = (α + β)^2 - 4 α β$

= $({11}/2)^4 -4(5/2) = {121}/4 - {40}/4 = {81}/4 = (9/2)^2$

∴ Difference of roots = (α - β) = $9/2$ = 4.5.

Answer: (c)

(b - 6) is a root of

$x^2$ - 6x + b = 0

⇒ $(b - 6)^2 - 6(b - 6) + b = 0$

$b^2$ + 36 - 12b - 6b + 36 + b = 0

$b^2$ - 17b + 72 = 0

$b^2$ - 9b - 8b + 72 = 0

b(b - 9) - 8(b - 9) = 0

(b - 8) (b - 9) = 0

⇒ b = 9 or b = 8

Maximum value of $b^2 = 9^2$ = 81

Answer: (b)

α + β = 6 and α β = 6

∴ $(α + β)^2 = 6^2$

⇒ $α^2 + β^2$ + 2 α β = 36

⇒ $α^2 + β^2$ = 36 - 2 (6) = 24

∴ $(α^3 + β^3) + (α^2 + β^2) + (α + β)$

= $(α + β) (α^2 + β^2 - α β) + (α^2 + β^2) + (α + β)$

= 6(24 - 6) + (24) + (6)

= 6(18) + 30 = 108 + 30 = 138

Answer: (d)

Given, $px^2$ + qx + r = 0

Let α and β are the roots of equation.

According to the question,

β = 2 α

Product of roots (α β) = $r/p = 2 α^2$

⇒ $α^2 = r/{2p}$

Sum of roots (α + β) = -$q/p$ = 3 α

⇒ α = ${- q}/{3p}$

We put the value pf α in equation (i),

Now,

⇒ $(q/{3p})^2 = r/{2p}$

$2q^2$ = 9pr

Answer: (a)

Given that,

$2x^2$ + 6x + 5y + 1 = 0.......(i)

and 2x + y + 3 = 0 ⇒ x = ${-y -3}/2$ .....(ii)

Now, putting the value of x from (ii) in equation (i),

$2({-3 - y}/2)^2 + 6({-3 -y}/2)$ + 5y + 1 = 0

⇒ ${9 + y^2 + 6y}/2 - {(18y + 6y)}/2$ + 5y + 1 = 0

⇒ $y^2$ + 10y - 7 = 0