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

Answer: (d)

Given,

x + y = 5

y + z = 10

z + x = 15

By solving equations (i), (ii) and (iii), then

x = 5, y = 0 and z = 10

∴ z > x > y

Answer: (b)

Let two positive integers be x and x + 1.

According to question,

⇒ $x^2 + (x + 1)^2$ = 761

$2x^2$ + 2x - 760 = 0

⇒ $x^2$ + x - 380 = 0

⇒ (x + 20) (x - 19) = 0

∴ x = 19 (∵ x ≠ - 20)

So, two consecutive integers are 19 and 20.

Answer: (a)

Let the common root be α, then

$α^2$ - k α - 21 = 0 ......(i)

and $α^2$ - 3kα + 35 = 0 .....(ii)

Solving by the rules of cross multiplication

${α^2}/{-35 - 63k} = {α}/{-21 - 35} = 1/{-3k + k}$

⇒${α^2}/{-98k} ={α}/{-56} = 1/{-2k}$

⇒${α^2}/{-98k} ={-1}/{2k}$

⇒$α^2$ = 49 and α = ${-56}/{-2k} ={28}/k$

Then, $({28}/k)^2 = 49$

⇒${28 × 28}/{49} = k^2$

⇒ 16 = $k^2$

∴ k = ± 4.

Answer: (d)

Let roots of equation are α and β

∴ α + β = ${a^2 + b^2}/{a^2 b^2} and α β = 1/{a^2 b^2}$

We know that

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

⇒ α - β = $√{{(a^2 - b^2)^2}/{(a^2 b^2)^2}} = {a^2 - b^2}/{a^2 b^2}$

On solving, we get α = $1/{b^2} and β = 1/{a^2}$

Answer: (b)

α + β = ${- (-1)}/1 = 1$

α . β = ${-1}/1$ = -1

Now, $(α + β)^2 = α^2 + β^2$ + 2 α β

⇒ $α^2 + β^2 = (1)^2 - 2 × (-1) = 3$

and $α^2 - β^2 = √{(α^2 + β^2)^2 - 4 α^2 β^2}$

= $√{9 - 4 (-1)^2} = √5$

α - β = $√{(α - β)^2 - 4 α β}$

= $√{1 - 4 (-1)} = √5$

Now, ${α^2 + β^2}/{(α^2 - β^2) (α - β)} = 3/{√5. √5} = 3/5$