# Free Quadratic Equations 02 Practice Test - 10th Grade

### Question 1

Find the value of k for which x2–4x+k=0 has coincident roots.

0

-2

4

-4

#### SOLUTION

Solution :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=16−4kThe roots of quadratic equation are co-incident only when D=0.

⇒16−4k=0

⇒k=4

### Question 2

The product of 2 consecutive natural numbers is 56. Solve this problem using completing the square method of quadratic equation. Write the equation obtained after completing the square and also write the smaller of the 2 numbers.

7, (x+12)2−2254=0

8, (x+12)2–2254=0

7, (x−12)2−2254 = 0

8, (x−12)2−2254=0

#### SOLUTION

Solution :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.

### Question 3

The nature of roots of x2−3x+2=0 will be:

Two distinct real roots.

Two equal real roots.

No real roots.

None of these.

#### SOLUTION

Solution :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.

### Question 4

During a practice match, a softball pitcher throws a ball whose height can be modeled by the equation h=−16t2+24t+1, where h = height in feet and t = time in seconds. How long does it take for the ball to reach a height of 6 feet?

2.2 and 3.8 secs

5.4 and 6.2 secs

0.25 and 1.25 secs

7 and 5 secs

#### SOLUTION

Solution :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.

### Question 5

One leg of a right angled triangle exceeds the other leg by 4 inches. The hypotenuse is 20 inches. Find the length of the shorter leg of the triangle.

12 inches

7 inches

20 inches

14 inches

#### SOLUTION

Solution :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 .

### Question 6

If the roots of equation x2+4x+c=0 are real then c≥4.

True

False

#### SOLUTION

Solution :B

Given, x2+4x+c=0

Value of discriminant, Δ=b2−4ac=42–4c=16−4cThe 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

### Question 7

Solve the following quadratic equation using quadratic formula .

9x2−9(a+b)x+(2a2+5ab+2b2)=0

The roots are 2a+b3 and a+2b3

The roots are 2a+b3 and a−2b3

The roots are 5a+b3 and a+2b3

The roots are 2a+b3 and a−2b4

#### SOLUTION

Solution :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

### Question 8

Determine the value of k so that the two roots of the equation x^{2} − kx + 36 = 0 are equal.

#### SOLUTION

Solution :A and B

We know that roots of the quadratic equation is equal when

D= 0

⇒b2−4ac=0

So, from the equation x^{2}− kx + 36 = 0

a = 1, b = -k and c = 36

⇒ k2−4(ac)=0

k2−4(36)=0

k2=144

∴ k = 12 or -12

### Question 9

11 years from now, the age of Peter will be half the square of the age he was 13 years ago. Calculate the current age of Peter.

21 years

7 years

27 years

25 years

#### SOLUTION

Solution :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

### Question 10

A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B alone to finish the work.

B alone can finish the work in 9 days

B alone can finish the work in 8 days

B alone can finish the work in 12 days

B alone can finish the work in 5 days

#### SOLUTION

Solution :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