Quadratic Equations Question Answers

Question 1

If one root of the quadratic equation $a x^{2}$ + bx + c = 0 is equal to the n^th power of the other root, then the value of $(a c^{n})^{\frac{1}{n + 1}}$ + $(a^{n} c)^{\frac{1}{n + 1}}$

A.b

B.-b

b^{\frac{1}{n + 1}}

D.-

b^{\frac{1}{n + 1}}

SOLUTION

Solution : B

Let $α$ , $α^{n}$ be two roots,

Then $α + α^{n}$ = – $\frac{b}{a}$ , $α α^{n}$ = $\frac{c}{a}$

Eliminating $α$ , we get ( $\frac{c}{a})^{\frac{1}{n + 1}}$ + ( $\frac{c}{a})^{\frac{n}{n + 1}}$ = – $\frac{b}{a}$

⇒ $a . a^{- \frac{1}{n + 1}}$ . $c^{\frac{1}{n + 1}}$ + $a . a^{- \frac{n}{n + 1}}$ . $c^{\frac{n}{n + 1}}$ = -b

or $(a^{n} c)^{\frac{1}{n + 1}}$ + $(a c^{n})^{\frac{1}{n + 1}}$ = -b

Question 2

If one root of the equation $a x^{2}$ + bx + c = 0 be n times the other root, then

n a^{2}

= bc

(n + 1)^{2}

n b^{2}

= ac

(n + 1)^{2}

n c^{2}

= ab

(n + 1)^{2}

n b^{3}

= ac

(n + 1)^{2}

SOLUTION

Solution : B

Let the roots be $α$ and n $α$

Sum of roots, α + nα = – $\frac{b}{a}$ ⇒ α = – $\frac{b}{a (n + 1)}$ ……(i)

and product, α.n.α = $\frac{c}{a}$ ⇒ $α^{2}$ = $\frac{c}{n a}$ ……(ii)

From (i) and (ii), we get

⇒ $[$ – $\frac{b}{a (n + 1)}$ $]^{2}$ = $\frac{c}{n a}$ ⇒ $\frac{b^{2}}{a^{2} (n + 1)^{2}}$ = $\frac{c}{n a}$

⇒ $n b^{2}$ = ac $(n + 1)^{2}$ .

Note : Students should remember this question as a fact.

Question 3

The coefficient of x in the equation $x^{2}$ + p $x$ + q = 0 was taken as 17 in place of 13, its roots were found to be -2 and -15, the roots of the original equation are

A.3, 10

B.- 3, – 10

C.- 5, – 18

D.18,5

SOLUTION

Solution : B

Let the equation (in written form) be $x^{2}$ + 17 $x$ + q = 0.
Roots are -2, -15.
So q = 30,
And correct equation is $x^{2}$ + 13 $x$ + 30 = 0.
Hence roots are -3, -10.

Question 4

The values of a for which( $a^{2} - 1$ ) $x^{2}$ + 2(a-1)x + 2 is positive for any x are

A.a ≥ 1

B.a ≤ 1

C.a > -3

D.a < -3 or a > 1

SOLUTION

Solution : D

We know that the expression a $x^{2}$ + b $x$ + c > 0 for all $x$ , if a > 0 and $b^{2}$ < 4ac

∴ $(a^{2} - 1)$ $x^{2}$ + 2(a-1) $x$ + 2 is positive for all $x$ , if $a^{2} - 1$ > 0 and 4 $(a - 1)^{2} - 8 (a^{2} - 1)$ < 0

⇒ $a^{2} - 1$ > 0 and -4(a – 1)(a + 3) < 0

⇒ $a^{2} - 1$ > 0 and (a – 1)(a + 3) > 0

⇒ $a^{2}$ > 1 and a < -3 or a > 1

⇒ a < -3 or a > 1

Question 5

If the roots of the equation q $x^{2}$ + p $x$ + q = 0 where p, q are real, are complex, then the roots of the equation $x^{2}$ – 4q $x$ + $p^{2}$ = 0 are

A.Real and unequal

B.Real and equal

C.Imaginary

D.nothing can be said in particular

SOLUTION

Solution : A

The given equations are

q $x^{2}$ + p $x$ + q = 0 ……..(i)

and $x^{2}$ – 4q $x$ + $p^{2}$ = 0 ……..(ii)

Roots of (i) are complex, therefore $p^{2} - 4 q^{2}$ < 0

Now discriminant of (ii) is

$16 q^{2} - 4 p^{2} = - 4 (p^{2} - 4 q^{2})$ > 0

Hence, roots are real and unequal.

Question 6

If $x$ is real and $k = \frac{x^{2} - x + 1}{x^{2} + x + 1}$ , then

\frac{1}{3} \leq k \leq 3

B.k ≥ 5

C.k ≤ 0

D.k

\leq

SOLUTION

Solution : A

From $k = \frac{x^{2} - x + 1}{x^{2} + x + 1}$

We have $x^{2} (k - 1) + x (k + 1) + k - 1 = 0$

As given, x is real ⇒ $(k + 1)^{2} - 4 (k - 1)^{2} \geq 0$

⇒ $3 k^{2} - 10 k + 3 \leq 0$

Which is possible only when the value of k lies between the roots of the equation $3 k^{2} - 10 k + 3 = 0$

That is, when $\frac{1}{3} \leq k \leq 3$ {Since roots are $\frac{1}{3}$ and 3}

Question 7

If $a x^{2} + b x + c = 0$ , then x =

\frac{b \pm \sqrt{b^{2} - 4 a c}}{2 a}

\frac{- b \pm \sqrt{b^{2} - a c}}{2 a}

\frac{2 c}{- b \pm \sqrt{b^{2} - 4 a c}}

D.None of these

SOLUTION

Solution : D

Try making perfect square the given equation by adding and subtracting $\frac{b^{2}}{4 a^{2}}$
on solving it we’ll find
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Question 8

If the sum of two of the roots of $x^{3} + p x^{2} + q x + r = 0$ is zero, then pq =

A.- r

B.r

C.2 r

D.- 2 r

SOLUTION

Solution : B

Given that, α + β = 0

α + β + γ = -p ⇒ γ = -p

Substituting γ = -p in the given equation

⇒ $- p^{3} + p^{3} - p q + r = 0$ ⇒ pq =r

Question 9

If α, β, γ are the roots of the equation $x^{3} + 4 x + 1 = 0$ ,then $(α + β)^{- 1} + (β + γ)^{- 1} + (γ + α)^{- 1} =$

A.2

B.3

C.4

D.5

SOLUTION

Solution : C

If $α, β$ , γ are the roots of the equation

$α + β + γ = 0, α β + β γ + γ α = 4, α β γ = - 1$

$t h e r e f o r e (α + β)^{- 1} + (β + γ)^{- 1} + (γ + α)^{- 1}$
= $\frac{1}{- γ} + \frac{1}{- α} + \frac{1}{- β}$
= $- (\frac{α β + β γ + γ α}{α β γ})$
= $- (\frac{4}{- 1}) = 4$

Question 10

If $α, β$ and $γ$ are the roots of $x^{3} + 8 = 0$ , then the equation whose roots are $α^{2}, β^{2} a n d γ^{2}$ is

x^{3} - 8 = 0

x^{3} - 16 = 0

x^{3} + 64 = 0

x^{3} - 64 = 0

SOLUTION

Solution : D

Let $y = x^{2}$ . then $x = \sqrt{y}$

∴ $x^{3} + 8 = 0$ ⇒ $y^{\frac{3}{2}} + 8 = 0$

⇒ $y^{3} = 64$ ⇒ $y^{3} - 64 = 0$

Thus the equation having roots $α^{2}, β^{2}$ and $γ^{2} i s x^{3} - 64 = 0$

Question 11

If two roots of the equation $x^{3} - 3 x + 2 = 0$ are same, then the roots will be

A.2, 2, 3

B.1, 1, -2

C.- 2, 3, 3

D.-2, -2, 1

SOLUTION

Solution : B

Given equation is $x^{3} - 3 x + 2 = 0$

⇒ $x^{2} (x - 1) + x (x - 1) - 2 (x - 1) = 0$

⇒ $(x - 1) (x^{2} + x - 2) = 0$ ⇒ $(x - 1) (x - 1) (x + 2) = 0$

Hence roots are 1, 1, -2

Question 12

One root of the following given equation $2 x^{5} - 14 x^{4} + 31 x^{3} - 64 x^{2} + 19 x + 130 = 0$ is

A.1

B.3

C.5

D.7

SOLUTION

Solution : C

We can solve this problem using options given –
On putting x = 5, we get

$2 (5)^{5} - 14 (5)^{4} + 31 (5)^{3} - 64 (5)^{2} + 19 (5) + 130 = 0$

Hence x = 5 satisfies the given equation.

Thus 5 is a root of the equation.
Other options will not satisfy the given equation so the answer is x = 5.

Question 13

If the roots of the equation $8 x^{3} - 14 x^{2} + 7 x - 1 = 0$ are in G.P., then the roots are

A.1,

\frac{1}{2}

\frac{1}{4}

B.2, 4, 8

C.3, 6, 12

D.1,2,-4

SOLUTION

Solution : A

Let the roots be $\frac{α}{β}, α, α β, β \neq 0$
Then the product of roots is $α^{3} = - \frac{- 1}{8} = \frac{1}{8}$
⇒ $α = \frac{1}{2}$ and hence β = $\frac{1}{2}$
(Since, sum of roots $= \frac{α}{β} + α + α β = 14 / 8)$ ,
so roots are $1, \frac{1}{2}$ , $\frac{1}{4}$

Alternate method: By inspecting through options, we get the numbers $1, \frac{1}{2}$ , $\frac{1}{4}$ satisfying the given equation.

Question 14

If $x^{2} + 2 a x + 10 - 3 a > 0$ for all x ∈ R, then

A.-5 < a < 2

B.a < -5

C.a > 5

D.2 < a < 5

SOLUTION

Solution : A

According to given condition,

$4 a^{2} - 4 (10 - 3 a) < 0$ ⇒ $a^{2} + 3 a - 10 < 0$

⇒ $(a + 5) (a - 2) < 0 \Rightarrow - 5 < a < 2.$

Question 15

The set of all real numbers x for which $x^{2} - | x + 2 | + x > 0$ , is

A. (-∞, -2)∪ (2, ∞)

B.(-∞, –

\sqrt{2}

)∪(

\sqrt{2}

, ∞)

C.(-∞, -1)∪(1, ∞)

D.(

\sqrt{2}

, ∞)

SOLUTION

Solution : B

Case 1: When x + 2 ≥ 0 i.e/ x ≥ -2,

Then given inequality becomes

$x^{2} - (x + 2) + x > 0$ ⇒ $x^{2} - 2 > 0$ ⇒ $| x | > \sqrt{2}$

⇒ therefore, in this case the part of the solution set is

[-2,- $\sqrt{2}$ ) ∪ ( $\sqrt{2}$ , ∞).

Case 2: When x + 2 < 0 i.e. x < -2,

Then given inequality becomes $x^{2} + (x + 2) + x > 0$

⇒ $x^{2} + 2 x + 2 > 0$ ⇒ $(x + 1)^{2} + 1 > 0$ , which is true for all real x

Hence, the part of the solution set in this case is (-∞, -2). Combining the two cases, the solution set is (-∞, -2)∪[-2, – $\sqrt{2}$ )∪( $\sqrt{2}, \infty) = (- \infty, - \sqrt{2}$ )∪( $\sqrt{2}$ , ∞)