Quantitative Aptitude
QUADRATIC EQUATIONS MCQs
Quadratic Equations, 10th And 12th Grade Quadratic Equations
Total Questions : 102
| Page 4 of 11 pages
Answer: Option C. -> 1, –5
:
C
Step 1:- x2+2(k+2)x+9=0, ⇒a=1,b=2(k+2),c=9
D=4(k+2)2–4(9)=4(k2+4k−5) [D=b2−4ac ]
Step 2:- The roots of quadratic equation are real and equal only when D=0
4(k2+4k−5)=0,⇒k2+4k−5=0
Step 3:- k2+4k−5=0
⇒k2+5k−k−5=0
⇒k(k+5)−1(k+5)=0
⇒(k−1)(k+5)=0
⇒k=1ork=−5
:
C
Step 1:- x2+2(k+2)x+9=0, ⇒a=1,b=2(k+2),c=9
D=4(k+2)2–4(9)=4(k2+4k−5) [D=b2−4ac ]
Step 2:- The roots of quadratic equation are real and equal only when D=0
4(k2+4k−5)=0,⇒k2+4k−5=0
Step 3:- k2+4k−5=0
⇒k2+5k−k−5=0
⇒k(k+5)−1(k+5)=0
⇒(k−1)(k+5)=0
⇒k=1ork=−5
Answer: Option A. -> 1 or 4
:
A
Step 1:- For, x2+2(k+2)x+9k=0, value of discriminant D=[2(k+2)]2–4(9k)=4(k2+4−5k)
Step 2:- The roots of quadratic equation are real and equal only when D=0
k2+4−5k=0
⇒k2−5k+4=0
⇒k2−k−4k+4=0
⇒k(k−1)−4(k−1)=0
⇒(k−1)(k−4)=0
Step 3:- k=4or1
:
A
Step 1:- For, x2+2(k+2)x+9k=0, value of discriminant D=[2(k+2)]2–4(9k)=4(k2+4−5k)
Step 2:- The roots of quadratic equation are real and equal only when D=0
k2+4−5k=0
⇒k2−5k+4=0
⇒k2−k−4k+4=0
⇒k(k−1)−4(k−1)=0
⇒(k−1)(k−4)=0
Step 3:- k=4or1
Answer: Option B. -> 3, (x−74)2−2516=0
:
B
2x2−7x+3=0
Dividing by the coefficient of x2, we get
x2−72x+32=0;a=1,b=72,c=32
Adding and subtracting the square of b2=74, (half of coefficient of x)
we get,
[x2−2(74)x+(74)2]−(74)2+32=0
The equation after completing the square is :
(x−74)2−2516=0
Taking square root, (x−74)=(±54)
Taking positive sign 54,x=3
Taking negative sign −54,x=12
:
B
2x2−7x+3=0
Dividing by the coefficient of x2, we get
x2−72x+32=0;a=1,b=72,c=32
Adding and subtracting the square of b2=74, (half of coefficient of x)
we get,
[x2−2(74)x+(74)2]−(74)2+32=0
The equation after completing the square is :
(x−74)2−2516=0
Taking square root, (x−74)=(±54)
Taking positive sign 54,x=3
Taking negative sign −54,x=12
Answer: Option A. -> -8
:
A
If one root of the equation 2x2+ax+6=0 is 3, then substitutingx=3 will satisfythe equation.
2(3)2+3a+6=0
⇒18+3a+6=0
⇒24+3a=0
⇒a=−8
:
A
If one root of the equation 2x2+ax+6=0 is 3, then substitutingx=3 will satisfythe equation.
2(3)2+3a+6=0
⇒18+3a+6=0
⇒24+3a=0
⇒a=−8
Answer: Option A. -> -1
:
A
Since, x = 2 is a root of the given equation 2x2+ax−6=0
→ 2(2)2+a×2−6=0
8+2a−6=0
2a=−2
a=−1
:
A
Since, x = 2 is a root of the given equation 2x2+ax−6=0
→ 2(2)2+a×2−6=0
8+2a−6=0
2a=−2
a=−1
Answer: Option C. -> ac
:
C
Let roots of x2−cx+d=0 beα,β then roots of x2−ax+b=0 beα,α
∴α +β = c,αβ = d,α +α =α, α2 = b
Hence 2(b+d)=2(α2+αβ)=2α(α+β)=ac
:
C
Let roots of x2−cx+d=0 beα,β then roots of x2−ax+b=0 beα,α
∴α +β = c,αβ = d,α +α =α, α2 = b
Hence 2(b+d)=2(α2+αβ)=2α(α+β)=ac
Answer: Option B. -> x2−4abx−(a2−b2)2=0
:
B
Sum of rootsα +β = -(a + b) andαβ = a2+b22
⇒(α+β)2=(a+b)2and(α−β)2=α2+β2−2αβ
=2ab−(a2+b2)=−(a−b)2
Now the required equation whose roots are
(α+β)2and(α−β)2
x2−{(α+β)2+(α−β)2}x+(α+β)2(α−β)2=0
⇒x2−{(a+b)2+(a−b)2}x+(a+b)2(a−b)2=0
⇒x2−4abx−(a2−b2)2=0
:
B
Sum of rootsα +β = -(a + b) andαβ = a2+b22
⇒(α+β)2=(a+b)2and(α−β)2=α2+β2−2αβ
=2ab−(a2+b2)=−(a−b)2
Now the required equation whose roots are
(α+β)2and(α−β)2
x2−{(α+β)2+(α−β)2}x+(α+β)2(α−β)2=0
⇒x2−{(a+b)2+(a−b)2}x+(a+b)2(a−b)2=0
⇒x2−4abx−(a2−b2)2=0
Answer: Option D. -> 4
:
D
Subtracting, we get 2hx = 56 or hx = 28
Putting in any, x2=49
∴ [28h]2=72⇒ h = 4(h > 0)
:
D
Subtracting, we get 2hx = 56 or hx = 28
Putting in any, x2=49
∴ [28h]2=72⇒ h = 4(h > 0)
Answer: Option C. -> 1-b
:
C
Given equation x2−a(x+1)−b=0
⇒ x2−ax−a−b⇒α +β =α,αβ = -(a + b)
Now(α+1)(β+1)=αβ+α+β+1=−(a+b)+a+1=1−b
:
C
Given equation x2−a(x+1)−b=0
⇒ x2−ax−a−b⇒α +β =α,αβ = -(a + b)
Now(α+1)(β+1)=αβ+α+β+1=−(a+b)+a+1=1−b
Answer: Option D. -> −23
:
D
Under condition −2λ=3 ⇒ λ=−23
:
D
Under condition −2λ=3 ⇒ λ=−23
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