Quantitative Aptitude
QUADRATIC EQUATIONS MCQs
Quadratic Equations, 10th And 12th Grade Quadratic Equations
Total Questions : 102
| Page 2 of 11 pages
Answer: Option B. -> 1, 1, -2
:
B
Given equation is x3−3x+2=0
⇒ x2(x−1)+x(x−1)−2(x−1)=0
⇒ (x−1)(x2+x−2)=0⇒ (x−1)(x−1)(x+2)=0
Hence roots are 1, 1, -2
:
B
Given equation is x3−3x+2=0
⇒ x2(x−1)+x(x−1)−2(x−1)=0
⇒ (x−1)(x2+x−2)=0⇒ (x−1)(x−1)(x+2)=0
Hence roots are 1, 1, -2
Answer: Option A. -> -5 < a < 2
:
A
According to given condition,
4a2−4(10−3a)<0 ⇒ a2+3a−10<0
⇒ (a+5)(a−2)<0⇒−5<a<2.
:
A
According to given condition,
4a2−4(10−3a)<0 ⇒ a2+3a−10<0
⇒ (a+5)(a−2)<0⇒−5<a<2.
Answer: Option A. -> 13≤k≤3
:
A
Fromk=x2−x+1x2+x+1
We have x2(k−1)+x(k+1)+k−1=0
As given, x is real ⇒ (k+1)2−4(k−1)2≥0
⇒ 3k2−10k+3≤0
Which is possible only when the value of k lies between the roots of the equation3k2−10k+3=0
That is, when 13≤k≤3 {Since roots are 13 and 3}
:
A
Fromk=x2−x+1x2+x+1
We have x2(k−1)+x(k+1)+k−1=0
As given, x is real ⇒ (k+1)2−4(k−1)2≥0
⇒ 3k2−10k+3≤0
Which is possible only when the value of k lies between the roots of the equation3k2−10k+3=0
That is, when 13≤k≤3 {Since roots are 13 and 3}
Answer: Option C. -> 3
:
C
The condition for common roots gives (bc−a2)2=(ca−b2)(ab−c2)On simplification gives a3+b3+c3=3abc
:
C
The condition for common roots gives (bc−a2)2=(ca−b2)(ab−c2)On simplification gives a3+b3+c3=3abc
Answer: Option B. -> False
:
B
Given, x2+4x+c=0
Value of discriminant, Δ=b2−4ac=42–4c=16−4c
The roots of quadratic equation are real thenΔ≥016–4c≥0−4c≥−164c≤16
(The inequality changes when the equation is multiplied by ′−ve′
∴c≤4
:
B
Given, x2+4x+c=0
Value of discriminant, Δ=b2−4ac=42–4c=16−4c
The roots of quadratic equation are real thenΔ≥016–4c≥0−4c≥−164c≤16
(The inequality changes when the equation is multiplied by ′−ve′
∴c≤4
Answer: Option C. -> 0.25 and 1.25 secs
:
C
Given Height = 6
⇒−16t2+24t+1=6
⇒16t2−24t+5=0
⇒16t2−4t−20t+5=0
⇒16t2−4t−20t+5=0
⇒4t(4t−1)−5(4t−1)=0
⇒(4t−1)(4t−5)=0
t=14,54or0.25,1.25
So, at time 0.25 secs and 1.25 secs, the ball will be at a height of 6 feet.
:
C
Given Height = 6
⇒−16t2+24t+1=6
⇒16t2−24t+5=0
⇒16t2−4t−20t+5=0
⇒16t2−4t−20t+5=0
⇒4t(4t−1)−5(4t−1)=0
⇒(4t−1)(4t−5)=0
t=14,54or0.25,1.25
So, at time 0.25 secs and 1.25 secs, the ball will be at a height of 6 feet.
Answer: Option A. -> 7, (x+12)2−2254=0
:
A
Let the smaller number bex. Then the larger numberwill be x+1.
Their product is x(x+1).
x(x+1)=56
x2+x−56=0
x2+x+(12)2–(12)2−56=0
(x+12)2−2254=0
x+12=152,−152
x=7,−8
The required number is 7 as -8 is not a natural number.
:
A
Let the smaller number bex. Then the larger numberwill be x+1.
Their product is x(x+1).
x(x+1)=56
x2+x−56=0
x2+x+(12)2–(12)2−56=0
(x+12)2−2254=0
x+12=152,−152
x=7,−8
The required number is 7 as -8 is not a natural number.
Answer: Option C. -> 4
:
C
On comparing x2–4x+k=0 with standard form ax2+bx+c=0, we get
a = 1, b = -4 and c = k
Now, discriminant, D = b2−4ac
⇒D=(−4)2–4(1)k=16−4k
The roots of quadratic equation are co-incident only when D=0.
⇒16−4k=0
⇒k=4
:
C
On comparing x2–4x+k=0 with standard form ax2+bx+c=0, we get
a = 1, b = -4 and c = k
Now, discriminant, D = b2−4ac
⇒D=(−4)2–4(1)k=16−4k
The roots of quadratic equation are co-incident only when D=0.
⇒16−4k=0
⇒k=4
Answer: Option A. -> Two distinct real roots.
:
A
We have, x2−3x+2=0
⇒D=b2−4ac
⇒(−3)2−4×1×2>0
⇒1>0
Since, D>0, roots are distinct and real.
:
A
We have, x2−3x+2=0
⇒D=b2−4ac
⇒(−3)2−4×1×2>0
⇒1>0
Since, D>0, roots are distinct and real.
Answer: Option A. -> 21 years
:
A
Let Peter's current age = x years
∴ Age 13 years ago= (x−13) years
Age 11 years later = (x+11) years
According to the given statement
11 years later , the age of Peter = Half the square of the age he was 13 years ago
⇒ (x+11) = (x−13)22
⇒ 2x+22=x2−26x+169
⇒ x2−28x+147=0
⇒ x2−(21+7)x+147=0
⇒x2−21x−7x+147=0
⇒ x(x−21)−7(x−21)=0
⇒ x=21orx=7
Since Peter's age can't be 7 years
∴ His current age = 21 years
:
A
Let Peter's current age = x years
∴ Age 13 years ago= (x−13) years
Age 11 years later = (x+11) years
According to the given statement
11 years later , the age of Peter = Half the square of the age he was 13 years ago
⇒ (x+11) = (x−13)22
⇒ 2x+22=x2−26x+169
⇒ x2−28x+147=0
⇒ x2−(21+7)x+147=0
⇒x2−21x−7x+147=0
⇒ x(x−21)−7(x−21)=0
⇒ x=21orx=7
Since Peter's age can't be 7 years
∴ His current age = 21 years
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