Quantitative Aptitude
QUADRATIC EQUATIONS MCQs
Quadratic Equations, 10th And 12th Grade Quadratic Equations
Total Questions : 102
| Page 6 of 11 pages
Answer: Option A. -> 1, 12, 14
:
A
Let the roots be αβ,α,αβ,β≠0
Then the product of roots is α3=−−18=18
⇒α=12 and henceβ = 12
(Since, sum of roots =αβ+α+αβ=14/8),
so roots are 1,12,14
Alternatemethod: By inspecting through options, we get the numbers1,12,14 satisfying the given equation.
:
A
Let the roots be αβ,α,αβ,β≠0
Then the product of roots is α3=−−18=18
⇒α=12 and henceβ = 12
(Since, sum of roots =αβ+α+αβ=14/8),
so roots are 1,12,14
Alternatemethod: By inspecting through options, we get the numbers1,12,14 satisfying the given equation.
Answer: Option D. -> x3−64=0
:
D
Let y=x2. then x=√y
∴ x3+8=0 ⇒ y32+8=0
⇒ y3=64⇒ y3−64=0
Thus the equation having roots α2,β2 and γ2isx3−64=0
:
D
Let y=x2. then x=√y
∴ x3+8=0 ⇒ y32+8=0
⇒ y3=64⇒ y3−64=0
Thus the equation having roots α2,β2 and γ2isx3−64=0
Answer: Option D. -> None of these
:
D
Try making perfect square the given equation by adding and subtracting b24a2
on solving it we'll find
x=−b±√b2−4ac2a
:
D
Try making perfect square the given equation by adding and subtracting b24a2
on solving it we'll find
x=−b±√b2−4ac2a
Answer: Option B. -> r
:
B
Given that,α +β= 0
α +β +γ = -p⇒ γ = -p
Substitutingγ = -p in the given equation
⇒ −p3+p3−pq+r=0⇒ pq =r
:
B
Given that,α +β= 0
α +β +γ = -p⇒ γ = -p
Substitutingγ = -p in the given equation
⇒ −p3+p3−pq+r=0⇒ pq =r
Answer: Option D. -> a < -3 or a > 1
:
D
We know that the expression ax2 + bx + c > 0 for all x, if a > 0 and b2 < 4ac
∴ (a2−1)x2 + 2(a-1)x + 2 is positive for allx, ifa2−1 > 0 and 4(a−1)2−8(a2−1) < 0
⇒ a2−1 > 0 and -4(a - 1)(a + 3) < 0
⇒ a2−1 > 0 and (a - 1)(a + 3) >0
⇒ a2 > 1 and a < -3 or a > 1
⇒ a < -3 or a > 1
:
D
We know that the expression ax2 + bx + c > 0 for all x, if a > 0 and b2 < 4ac
∴ (a2−1)x2 + 2(a-1)x + 2 is positive for allx, ifa2−1 > 0 and 4(a−1)2−8(a2−1) < 0
⇒ a2−1 > 0 and -4(a - 1)(a + 3) < 0
⇒ a2−1 > 0 and (a - 1)(a + 3) >0
⇒ a2 > 1 and a < -3 or a > 1
⇒ a < -3 or a > 1
Answer: Option A. -> Real and unequal
:
A
The given equations are
qx2 + px + q = 0 ........(i)
andx2 - 4qx + p2 = 0 ........(ii)
Roots of (i) are complex, therefore p2−4q2 < 0
Now discriminant of (ii) is
16q2−4p2=−4(p2−4q2) > 0
Hence, roots are realand unequal.
:
A
The given equations are
qx2 + px + q = 0 ........(i)
andx2 - 4qx + p2 = 0 ........(ii)
Roots of (i) are complex, therefore p2−4q2 < 0
Now discriminant of (ii) is
16q2−4p2=−4(p2−4q2) > 0
Hence, roots are realand unequal.
Answer: Option B. -> nb2 = ac(n+1)2
:
B
Let the roots be α and nα
Sum of roots,α + nα = -ba⇒α = -ba(n+1) ......(i)
and product,α.n.α = ca⇒ α2 = cna ......(ii)
From (i) and (ii), we get
⇒ [ -ba(n+1) ]2 = cna⇒ b2a2(n+1)2 = cna
⇒ nb2 = ac(n+1)2.
Note : Students should remember this question as a fact.
:
B
Let the roots be α and nα
Sum of roots,α + nα = -ba⇒α = -ba(n+1) ......(i)
and product,α.n.α = ca⇒ α2 = cna ......(ii)
From (i) and (ii), we get
⇒ [ -ba(n+1) ]2 = cna⇒ b2a2(n+1)2 = cna
⇒ nb2 = ac(n+1)2.
Note : Students should remember this question as a fact.
Answer: Option B. -> -b
:
B
Let α, αn be two roots,
Then α+αn = -ba, ααn =ca
Eliminating α, we get(ca)1n+1 +(ca)nn+1 = -ba
⇒ a.a−1n+1.c1n+1 +a.a−nn+1.cnn+1 = -b
or(anc)1n+1 + (acn)1n+1 = -b
:
B
Let α, αn be two roots,
Then α+αn = -ba, ααn =ca
Eliminating α, we get(ca)1n+1 +(ca)nn+1 = -ba
⇒ a.a−1n+1.c1n+1 +a.a−nn+1.cnn+1 = -b
or(anc)1n+1 + (acn)1n+1 = -b
Answer: Option C. -> 4
:
C
If α,β,γ are the roots of the equation
α+β+γ=0,αβ+βγ+γα=4,αβγ=−1
therefore(α+β)−1+(β+γ)−1+(γ+α)−1
=1−γ+1−α+1−β
=−(αβ+βγ+γααβγ)
=−(4−1)=4
:
C
If α,β,γ are the roots of the equation
α+β+γ=0,αβ+βγ+γα=4,αβγ=−1
therefore(α+β)−1+(β+γ)−1+(γ+α)−1
=1−γ+1−α+1−β
=−(αβ+βγ+γααβγ)
=−(4−1)=4
Answer: Option D. -> 4
:
D
Given|x|2−3|x|+2=0
Here we consider two cases viz.x < 0 and x > 0
Case 1: x < 0 This gives x2+3x+2=0
⇒ (x+2)(x+1)=0⇒ x=−2,−1
Also x=−1,−2 satisfy x<0, so x=−1,−2 is solution in this case
Case 2: x > 0. This gives x2−3x+2=0
⇒ (x-2)(x-1) = 0 ⇒x =2,1. so, x=2,1 is solution in this case. Hence the number5 of solutions are four i.e. x = -1, -2, 1, 2
Aliter : |x|2−3|x|+2=0
⇒ (|x|-1)(|x|-2)=0
⇒ |x| = 1 and |x| = 2 ⇒x =±1, x =±2.
:
D
Given|x|2−3|x|+2=0
Here we consider two cases viz.x < 0 and x > 0
Case 1: x < 0 This gives x2+3x+2=0
⇒ (x+2)(x+1)=0⇒ x=−2,−1
Also x=−1,−2 satisfy x<0, so x=−1,−2 is solution in this case
Case 2: x > 0. This gives x2−3x+2=0
⇒ (x-2)(x-1) = 0 ⇒x =2,1. so, x=2,1 is solution in this case. Hence the number5 of solutions are four i.e. x = -1, -2, 1, 2
Aliter : |x|2−3|x|+2=0
⇒ (|x|-1)(|x|-2)=0
⇒ |x| = 1 and |x| = 2 ⇒x =±1, x =±2.
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