Quantitative Aptitude
QUADRATIC EQUATIONS MCQs
Quadratic Equations, 10th And 12th Grade Quadratic Equations
Total Questions : 102
| Page 1 of 11 pages
Answer: Option C. -> 5
:
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.
:
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.
Answer: Option A. -> The roots are 2a+b3 and a+2b3
:
A
We have,
9x2−9(a+b)x+(2a2+5ab+2b2)=0
Comparing this equation with Ax2+Bx+C=0, we have
A=9,B=−9(a+b)andC=2a2+5ab+2b2
∴D=B2−4AC
⇒D=81(a+b)2−36(2a2+5ab+2b2)
⇒D=81(a2+b2+2ab)−(72a2+180ab+72b2)
⇒D=9a2+9b2−18ab
⇒D=9(a2+b2−2ab)
⇒D=9(a−b)2≥0
⇒D≥0
So, the roots of the given equation are real and are given by
α=−B+√D2A=9(a+b)+3(a−b)18=12a+6b18=2a+b3
and, β=−B−√D2A=9(a+b)−3(a−b)18=6a+12b18=a+2b3
:
A
We have,
9x2−9(a+b)x+(2a2+5ab+2b2)=0
Comparing this equation with Ax2+Bx+C=0, we have
A=9,B=−9(a+b)andC=2a2+5ab+2b2
∴D=B2−4AC
⇒D=81(a+b)2−36(2a2+5ab+2b2)
⇒D=81(a2+b2+2ab)−(72a2+180ab+72b2)
⇒D=9a2+9b2−18ab
⇒D=9(a2+b2−2ab)
⇒D=9(a−b)2≥0
⇒D≥0
So, the roots of the given equation are real and are given by
α=−B+√D2A=9(a+b)+3(a−b)18=12a+6b18=2a+b3
and, β=−B−√D2A=9(a+b)−3(a−b)18=6a+12b18=a+2b3
Answer: Option A. -> 12 inches
:
A
Let thelength of the legs of the triangle be x and x+4inches.
Applying Pythagoras' theorem,
x2+(x+4)2=(20)2x2+x2+8x+16=4002x2+8x−384=0x2+4x−192=0
Solving the quadratic equation,x2+16x−12x−192=0x(x+16)−12(x+16)=0(x+16)(x−12)=0∴x=−16&12
Length can't be negative, hence
x=12inches
The shortest side(leg) is 12 .
:
A
Let thelength of the legs of the triangle be x and x+4inches.
Applying Pythagoras' theorem,
x2+(x+4)2=(20)2x2+x2+8x+16=4002x2+8x−384=0x2+4x−192=0
Solving the quadratic equation,x2+16x−12x−192=0x(x+16)−12(x+16)=0(x+16)(x−12)=0∴x=−16&12
Length can't be negative, hence
x=12inches
The shortest side(leg) is 12 .
Answer: Option C. -> m=−16
:
C
Given: x=23 is a solution of the quadratic equation 7x2+mx−3=0.
Substitutingx=23 in the given quadratic equation we get
⇒ 7(23)2+m(23)−3=0
⇒ 7(49)+m(23)−3=0
⇒ 289+2m3−3=0
⇒ 28+6m−27=0
⇒ m=−16
:
C
Given: x=23 is a solution of the quadratic equation 7x2+mx−3=0.
Substitutingx=23 in the given quadratic equation we get
⇒ 7(23)2+m(23)−3=0
⇒ 7(49)+m(23)−3=0
⇒ 289+2m3−3=0
⇒ 28+6m−27=0
⇒ m=−16
Answer: Option D. -> ± 2
:
D
Given, the roots of quadratic equation 9x2+6kx+4=0are real and equal.
Discriminant,Δ=b2−4acΔ=36k2–4(4)(9)=36k2–4(36)
The roots of quadratic equation are real and equal thenΔ=036k2–4(36)=0k2−4=0k2=4k=√4∴k=±2
:
D
Given, the roots of quadratic equation 9x2+6kx+4=0are real and equal.
Discriminant,Δ=b2−4acΔ=36k2–4(4)(9)=36k2–4(36)
The roots of quadratic equation are real and equal thenΔ=036k2–4(36)=0k2−4=0k2=4k=√4∴k=±2
Answer: Option B. -> m
:
B
Step 1 :For, x2+2x+m=0,
⇒a=1,b=2,c=m
We know that
D=b2–4ac
⇒D=(2)2–4m
=4−4m
Step 2 :The roots of a quadratic equation are real and distinct only when D>0
Step 3 : 4−4m>0
4>4m
m<1
:
B
Step 1 :For, x2+2x+m=0,
⇒a=1,b=2,c=m
We know that
D=b2–4ac
⇒D=(2)2–4m
=4−4m
Step 2 :The roots of a quadratic equation are real and distinct only when D>0
Step 3 : 4−4m>0
4>4m
m<1
Answer: Option A. -> 400 km/hr
:
A
Let the speed of the train be x km/hr.
Time=DistanceSpeed
∴ Time takento cover 1200 km = 1200x
Reduced speed = (x−40) km/hr
∴ Time takento cover 1200 km in reduced speed = 1200x−40
Relation between the time taken to cover 1200 km with speed x and with speed (x - 40) km/hr is as given below
1200x−40−1200x=13 (∵20minutes=13hours)
⇒ 48000x(x−40)=13
⇒ 144000=x2−40x
⇒ x2−40x−144000=0
⇒ x2−400x+360x−144000=0
⇒ x(x−400)+360(x−400)=0
⇒ (x+360)(x−400)=0
⇒ x=−360orx=400
Since speed cannot be negative, we get x = 400 km/hr.
∴ The speed of the train is 400 km/hr.
:
A
Let the speed of the train be x km/hr.
Time=DistanceSpeed
∴ Time takento cover 1200 km = 1200x
Reduced speed = (x−40) km/hr
∴ Time takento cover 1200 km in reduced speed = 1200x−40
Relation between the time taken to cover 1200 km with speed x and with speed (x - 40) km/hr is as given below
1200x−40−1200x=13 (∵20minutes=13hours)
⇒ 48000x(x−40)=13
⇒ 144000=x2−40x
⇒ x2−40x−144000=0
⇒ x2−400x+360x−144000=0
⇒ x(x−400)+360(x−400)=0
⇒ (x+360)(x−400)=0
⇒ x=−360orx=400
Since speed cannot be negative, we get x = 400 km/hr.
∴ The speed of the train is 400 km/hr.
Answer: Option D. -> imaginary
:
D
Step 1:For 2x2–6x+8=0, value of discriminant
D=(−6)2–4(2)(8)=36−64=−28
Step 2:Since D<0, roots are imaginary.
:
D
Step 1:For 2x2–6x+8=0, value of discriminant
D=(−6)2–4(2)(8)=36−64=−28
Step 2:Since D<0, roots are imaginary.
Answer: Option A. -> −37
:
A
Given equation 4x2+3x+7=0, ∴
α+β=−34andαβ=74
Now 1α+1β=α+βαβ=−3474=−34×47=−37
:
A
Given equation 4x2+3x+7=0, ∴
α+β=−34andαβ=74
Now 1α+1β=α+βαβ=−3474=−34×47=−37
Answer: Option B. -> - 3, - 10
:
B
Let the equation (inwritten form) bex2 + 17x + q = 0.
Roots are -2, -15.
So q = 30,
And correct equation isx2 + 13x + 30= 0.
Hence roots are -3, -10.
:
B
Let the equation (inwritten form) bex2 + 17x + q = 0.
Roots are -2, -15.
So q = 30,
And correct equation isx2 + 13x + 30= 0.
Hence roots are -3, -10.
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