10th Grade > Mathematics
QUADRATIC EQUATIONS MCQs
:
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
:
A
Let the smaller number be x. Then the larger number will 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
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
Let the length of the legs of the triangle be x and x+4 inches.
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=12 inches
The shortest side(leg) is 12 .
:
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=14,54 or 0.25, 1.25
So, at time 0.25 secs and 1.25 secs, the ball will be at a height of 6 feet.
:
B
Given, x2+4x+c=0
Value of discriminant, Δ=b2−4ac=42–4c=16−4c
The roots of quadratic equation are real then Δ≥016–4c≥0−4c≥−164c≤16
(The inequality changes when the equation is multiplied by ′−ve ′
∴c≤4
:
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) and C=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 and B
We know that roots of the quadratic equation is equal when D = 0
⇒b2−4ac=0
So, from the equation x2 − kx + 36 = 0
a = 1, b = -k and c = 36
⇒ k2−4(ac)=0
k2−4(36)=0
k2=144
∴ k = 12 or -12
:
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=21 or x=7
Since Peter's age can't be 7 years
∴ His current age = 21 years
:
C
Suppose B alone can takesx days to finish the work.
Then A alone can finish it in (x−6) days.
B's 1 day's work = 1x.
A's 1 day's work = 1x−6.
(A+B)'s 1 day's work = 14
∴1x+1x−6=14
⇒x−6+xx(x−6)=14
⇒8x−24=x2−6x
⇒x2−14x+24=0
⇒x2−12x−2x+24=0
⇒(x−12)(x−2)=0
⇒x=12 or x=2
But x cannot be less than 6.
⇒x=12
Hence, B alone can finish the work in 12 days