Quadratic Equations(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

QUADRATIC EQUATIONS MCQs

Quadratic Equations, 10th And 12th Grade Quadratic Equations

Total Questions : 102 | Page 5 of 11 pages

Question 41. If the roots of the equation

a x^{2} + b x + c = 0

α a n d β

,, then the roots of the equation

c x^{2} + b x + a = 0

are

−α,−β
α,1β
1α, 1β
None of these

Discuss Question

Answer: Option C. -> 1α, 1β
:
C

α, β

are roots of

a x^{2} + b x + c = 0

\Rightarrow α + β

- \frac{b}{a} a n d α β = \frac{c}{a}

Let the roots of

c x^{2} + b x + a = 0

α

β

', then

α

' +

β

= - \frac{b}{c} a n d α β = \frac{a}{c}

but

\frac{α + β}{α β} = \frac{- \frac{b}{a}}{\frac{c}{a}} = \frac{- b}{c} \frac{1}{α} + \frac{1}{β} = α^{'} + β^{'}

Hence

α^{'} = \frac{1}{α} a n d β^{'} = \frac{1}{β}

Question 42. If α and β are the roots of the equation

x^{2} + 6 x + λ = 0

and

3 α + 2 β = - 20

, then λ =

-8
-16
16
8

Discuss Question

Answer: Option B. -> -16
:
B
α +β = -6 .....(i)
αβ =λ ......(ii)
and given 3α + 2β = -20 ......(iii)
Solving (i) and (iii), we getβ = 2,α = -8
Substituting these values in (ii), we getλ = -16

Question 43. If α and β are the roots of the equation

2 x^{2} - 3 x + 4 = 0

, then the equation whose roots are

α^{2}

and

β^{2}

4x2+7x+16=0
4x2+7x+6=0
4x2+7x+1=0
4x2−7x+16=0

Discuss Question

Answer: Option A. -> 4x2+7x+16=0
:
A
α +β =

\frac{3}{2}

andαβ = 2

α^{2} + β^{2} = (α + β)^{2} - 2 α β = \frac{9}{4} - 4 = - \frac{7}{4}

hence required equation

x^{2} - (α^{2} + β^{2}) x + α^{2} β^{2} = 0

⇒

x^{2} + \frac{7}{4} x + 4 = 0

⇒

4 x^{2} + 7 x + 16 = 0

Question 44. If

2 + i \sqrt{3}

is a root of the equation

x^{2} + p x + q = 0

, where p and q are real, then (p, q) =

(-4,7)
(4,7)
(4,7)
(-4,-7)

Discuss Question

Answer: Option A. -> (-4,7)
:
A
Since

2 + i \sqrt{3}

is a root, therefore

2 - i \sqrt{3}

will be other root. Now sum of the roots = 4 = -p and product of roots = 7 = q. Hence (p, q) = (-4, 7).

Question 45. If one root of

5 x^{2} + 13 x + k = 0

is reciprocal of the other, then k=

Discuss Question

Answer: Option B. -> 5
:
B
Let first root

= α

and second root

= \frac{1}{α}

Product of roots =1

\Rightarrow \frac{k}{5} = 1 \Rightarrow k = 5

Question 46. If a root of the equation

a x^{2} + b x + c = 0

be reciprocal of the equation then

a^{'} x^{2} + b^{'} x + c^{'} = 0

, then

(cc′−aa′)2=(ba′−cb′)(ab′−bc′).
(bb′−aa′)2=(ca′−bc′)(ab′−bc′).
(cc′−aa′)2=(ba′−cb′)(ab′−bc′).
(cc′+aa′)2=(ba′−cb′)(ab′−bc′).

Discuss Question

Answer: Option A. -> (cc′−aa′)2=(ba′−cb′)(ab′−bc′).
:
A
Letα be a root of first equation, then

\frac{1}{α}

be a root of second equation.
therefore

a α^{2} + b α + c = 0

and

a^{'} \frac{1}{α^{2}} + b^{'} \frac{1}{α} + c^{'} = 0

c^{'} α^{2} + b^{'} α + d = 0

Hence

\frac{α^{2}}{b a^{'} - b^{'} c} = \frac{α}{c c^{'} - a a^{'}} = \frac{1}{a b^{'} - b c^{'}}

(c c^{'} - a a^{'})^{2} = (b a^{'} - c b^{'}) (a b^{'} - b c^{'}) .

Question 47. If the ratio of the roots of the equation

a x^{2} + b x + c = 0

p : q

, then

pqb2+(p+q)2ac=0
pqb2−(p+q)2ac=0
pqa2−(p+q)2bc=0
pqb2−(p−q)2ac=0

Discuss Question

Answer: Option B. -> pqb2−(p+q)2ac=0
:
B
Let pα,qα be the roots of the given equation

a x^{2} + b x + c = 0

Then pα +qα =

- \frac{b}{a} a n d p α . q α = \frac{c}{a}

From first relation,α =

- \frac{b}{a (p + q)}

Substituting this value ofα in second relation, we get

\frac{b^{2}}{a^{2} (p + q)^{2}} \times p q = \frac{c}{a}

⇒

b^{2} p q - a c (p + q)^{2} = 0

Note : Students should remember this question as a fact.

Question 48. If

α, β

are the roots of the equation

a x^{2} + b x + c = 0

then the equation whose roots are

α + \frac{1}{β}

and

β + \frac{1}{α}

, is

acx2+(a+c)bx+(a+c)2=0
abx2+(a+c)bx+(a+c)2=0
acx2+(a+b)cx+(a+c)2=0
None of these

Discuss Question

Answer: Option A. -> acx2+(a+c)bx+(a+c)2=0
:
A

H e r e α + β =

- \frac{b}{a}

a n d α β =

\frac{c}{a}

If roots are

α + \frac{1}{β}, β + \frac{1}{α}

, then sum of roots are

= (α + \frac{1}{β}) + (β + \frac{1}{a l p h a}) = (α + β) + \frac{α + β}{α β} = - \frac{b}{a c} (a + c)

and product

= (α + \frac{1}{β}) (β + \frac{1}{α})

= α β + 1 + 1 + \frac{1}{α β} = 2 + \frac{c}{a} + \frac{a}{c}

= \frac{2 a c + c^{2} + a^{2}}{a c} = \frac{(a + c)^{2}}{a c}

Hence required equation is given by

x^{2} + \frac{b}{a c} (a + c) x + \frac{(a + c)^{2}}{a c} = 0

\Rightarrow

a c x^{2} + (a + c) b x + (a + c)^{2} = 0

Trick : Let a = 1, b = -3, c = 2,then

α

= 1,

β

= 2

∴ α + \frac{1}{β} = \frac{3}{2} a n d β + \frac{1}{α} = 3

∴

required equation must be

(x - 3) (2 x - 3) = 0

i.e.

2 x^{2} - 9 x + 9 = 0

Here (a) gives this equation on putting
a = 1, b = -3, c = 2

Question 49. If every pair of the equations

x^{2} + p x + q r = 0, x^{2} + q x + r p = 0, x^{2} + r x + p q = 0

have a common root, then the sum of three common roots is

−(p+q+r)2
−p+q+r2
-(p+q+r)
-p+q+r

Discuss Question

Answer: Option A. -> −(p+q+r)2
:
A
Let the roots beα,β;β,γ andγ,α respectively.
∴α +β = -p,β +γ= -q,γ +α= -r
Adding all, we get∑α =

\frac{- (p + q + r)}{2}

etc.

Question 50. The set of all real numbers x for which

x^{2} - | x + 2 | + x > 0

, is

(-∞, -2)∪ (2, ∞)
(-∞, -√2)∪(√2, ∞)
(-∞, -1)∪(1, ∞)
(√2, ∞)

Discuss Question

Answer: Option B. -> (-∞, -√2)∪(√2, ∞)
:
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, -