Quadratic Equations Test

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Quadratic Equations Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

Only one of the roots of ax² + bx + c = 0, a ≠ 0, is zero, if

A
c = 0

B
c = 0, b ≠ 0

C
b = 0, c = 0

D
b ≠ 0, c ≠ 0

Solution

When one root is zero and the other is non-zero, then
a(0)² + b(0) + c = 0
i.e. c = 0
Now, b ≠ 0 because if b = 0, then both the roots will be equal to 0.
As the equation will become
ax²= 0
Or x = 0, 0 (not possible, as both the roots are zero)
i.e. c = 0 and b ≠ 0
Question 2
1 / -0

Find the value of .

A
-2 -

B
¹/₄

C
-2 +

D
Cannot be determined

Solution

Let x =
Now, we can write this as .
⇒ x² + 4x - 1 = 0
⇒ x =
= = -2 ±
So, the roots are -2 - and -2 +.
But -2 - can't be the answer since it is a negative root.
⇒ x = -2 +
Question 3
1 / -0

For what values of c does the equation 2x² + 3x + c = 0 have imaginary roots?

A
c >

B
c >

C
c >

D
None of these

Solution

The equation has imaginary roots when b² - 4ac < 0.
Substituting a = 2 and b = 3 in b² - 4ac < 0,
3² - 4(2)c < 0
Or 9 - 8c < 0
Or 8c > 9
Or c > Thus, the roots are imaginary when c > .
Question 4
1 / -0

If the equation 9x² + 6kx + 4 = 0, k > 0, has equal roots, then both the roots are equal to

A

B

C

D

Solution

Given: 9x² + 6kx + 4 = 0 has equal roots.
∴ Determinant, D = 0 or b² - 4ac = 0 or b² = 4ac
36k² = 4 × 4 × 9
36k² = 36 × 4
k² = 4
k = ±2
But k > 0
∴ k = 2
∴ The equation becomes:
9x² + 12x + 4 = 0
9x² + 6x + 6x + 4 = 0
3x(3x + 2) + 2(3x + 2) = 0
(3x + 2)(3x + 2) = 0
x =
Question 5
1 / -0

The equation x - = 2 - has

A
infinitely many roots

B
two roots

C
one root

D
no root

Solution

x - = 2 -
It is to be noted that x - 1 cannot be 0 or x cannot be equal to 1.
So, multiplying both sides of the equation with (x - 1), we get the following equation.
x(x - 1) - 2 = 2(x - 1) - 2
Or x(x - 1) = 2(x - 1)
As x cannot be equal to 1, the only possible solution is x = 2.
Thus, the given equation has only one root.
Hence, answer option 3 is correct.
Question 6
1 / -0

If and are the roots of the quadratic equation ax² + bx + c = 0, then the value of is

A

B

C

D

Solution

= , =
+ = =
=
= =
=
Question 7
1 / -0

It is known that y² - 11y + 4k = 0 has two distinct roots. It is also known that one factor is (y - k) and k is greater than zero. What is the difference between the sum and products of the roots?

A
28

B
17

C
11

D
7

Solution

Since (y - k) is a factor of the quadratic equation, so substitute y = k in the equation.
So the equation y² - 11y + 4k = 0 transforms to k² - 11k + 4k = 0
Hence, k² = 7k and k = 7.
Now, substitute k = 7 in the equation y² - 11y + 4k = 0
The equation becomes y² - 11y + 28 = 0, the roots of which are 7 and 4.
Sum of the roots = 11;
Product of roots = 28
So, required difference = 28 - 11 = 17
Question 8
1 / -0

For what value of m will the equation (m + 1)x² + 2(m + 3)x + m + 8 = 0 have equal roots?

A
-1/2

B
1/3

C
-2

D
1

Solution

For equal roots, D = 0 or b² - 4ac = 0
4(m + 3)² - 4(m + 1)(m + 8) = 0
(m + 3)² - (m + 1)(m + 8) = 0
m² + 9 + 6m - m² - 9m - 8 = 0
-3m + 1 = 0
m = 1/3
Question 9
1 / -0

The values of a for which the expression (a² - 1)x² + 2 (a - 1)x + 2 is positive for any x are:

A
a ≥ 1

B
a ≤ 1

C
a > -3 or a ≤ 1

D
a < -3 or a ≥ 1

Solution

We know that the expression ax² + bx + c > 0 for all x, if a > 0 and b² < 4ac.
1(a² - 1)x² + 2(a - 1)x + 2 is positive for all x.
If a² - 1 > 0 and 4(a - 1)² - 8(a² - 1) < 0,
a² - 1 > 0 and -4(a - 1)(a + 3) < 0
⇒ a² - 1 > 0 and (a - 1)(a + 3) > 0
⇒ a < -1 or a > 1 and a < -3 or a > 1
⇒ a < -3 or a > 1
But at a = 1, given expression is always positive for any x
Therefore a < -3 or a ≥ 1
Question 10
1 / -0

If -4 is a root of the quadratic equation x² + px - 4 = 0 and the quadratic equation x² + px + k = 0 has equal roots, find the value of k.

A

B

C

D

Solution

x² + px - 4 = 0
(-4)² -4p - 4 = 0
16 - 4p - 4 = 0
4p = 12
p = 3
Now, x² + px + k = 0 has equal roots.
p² - 4k = 0
Question 11
1 / -0

If the equations ax² + bx + c = 0 and x² + 2x + 3 = 0 have common roots, then find the ratio a : b : c.

A
2 : 1 : 3

B
3 : 2 : 1

C
3 : 1 : 2

D
1 : 2 : 3

Solution

The equation x² + 2x + 3 = 0 has non-real roots, which can be equal to the roots of the equation ax² + bx + c = 0.
i.e. Both the roots are identical.
So,
∴ a = 1k, b = 2k and c = 3k
Ratio a : b : c = 1 : 2 : 3
Question 12
1 / -0

Solve for x:

A

B

C

D
None of these

Solution

Let (x + 1)/x = y
So, the equation becomes 1/y + y = 34/15 or 15y² - 34y + 15 = 0

15y² - 25y - 9y + 15 = 0

5y( 3y - 5) - 3(3y -5) = 0

(3y -5)( 5y -3) = 0

y = 3/5 or y = 5/3

(x +1) /x = 3/5

5x + 5 = 3x

2x = -5

x = -5/2

(x +1) /x = 5/3

3x +3 = 5x

x = 3/2 so, x = - 5/2 or 3/2.
Hence, option (3) is correct.
Question 13
1 / -0

If x = , then

A
bx² + ax + b = 0

B
bx² + ax – b = 0

C
bx² – ax – b = 0

D
bx² – ax + b = 0

Solution
Question 14
1 / -0

^{_If_and_{are the roots of the equation ax}2}^{_{+ bx + c = 0, then the equation whose roots are}}^_is

A
acx² - bx - 1 = 0

B
acx² – bx + 1 = 0

C
acx² + bx + 1 = 0

D
acx² + bx – 1 = 0

Solution

As α is a root of the equation ax² + bx + c = 0; aα² + bα + c = 0.
Or (aα + b)α = –c
Or … (1)
Similarly, … (2)
S = Sum of roots = = = (∵α + = –b/a)
P = Product of roots = = =
The equation is:
x² – Sx + P = 0
x² – x + = 0
acx² – bx + 1 = 0
Question 15
1 / -0

For what values of k will the quadratic equation x² + 4kx + k² - k + 2 = 0 have equal roots?

A

B

C

D

Solution

x² + 4kx + k² - k + 2 = 0
Here, a = 1, b = 4k and c = k² - k + 2
D = b² - 4ac
= (4k)² - 4(1)(k² - k + 2)
= 16k² - 4k² + 4k - 8
= 12k² + 4k - 8
Roots are equal.
∴ D = 0
So, 4(3k² + k - 2) = 0
3k² + 3k - 2k - 2 = 0
(3k - 2) (k + 1) = 0
k = and k = -1
Question 16
1 / -0

Each side of a square measures 4 cm more than each side of another square. If the sum of their areas is 400 sq. cm, then what is the area of the larger square?

A
144 cm²

B
169 cm²

C
196 cm²

D
256 cm²

Solution

Let each side of the smaller square measure x cm.
Then, length of each side of the larger square = (x + 4) cm
According to problem:
(x + 4)² + x² = 400 {Area of square = Side × Side}
x² + 8x + 16 + x² = 400
Or 2x² + 8x - 384 = 0
x² + 4x - 192 = 0
Or (x + 16)(x - 12) = 0
x = 12
So, length of each side of the larger square = (12 + 4) cm = 16 cm
Area = (16)² cm²= 256 cm²
Question 17
1 / -0

B's age is the square of the age of A. After 5 years, B will be 3 times as old as A. What is the difference between their ages?

A
5 years

B
10 years

C
15 years

D
20 years

Solution

Let the age of A be x.
So, B's age = x²
After 5 years, x² + 5 = 3(x + 5)
x² + 5 – 3x – 15 = 0
x² – 3x – 10 = 0 x = -2, 5
x is +ve, so x = 5
So, A's age = 5 years
B's age = 25 years
Difference = (25 – 5) years = 20 years
Thus, answer option 4 is correct.
Question 18
1 / -0

A motor boat takes 12 hours to go upstream and come back again to the starting point. The total distance travelled during the whole journey is 70 km. The speed of motor boat in still water, if the current of a stream runs at 1 km/hr, is

A
6 km/hr

B
7 km/hr

C
8 km/hr

D
8.5 km/hr

Solution

When going downstream, the speed of stream will get added to the speed of boat and when going upstream, it will be subtracted.
Let the speed of boat in still water be x km/hr.
Distance travelled upstream = Distance travelled downstream = 70/2 = 35 km
According to the question,

or, 35(2x) = 12(− 1)
or, 12x² - 70x - 12 = 0or, 12x² - 72x + 2x - 12 = 0
or, 12x(x - 6) + 2(x - 6) = 0
or, (x - 6)(12x + 2) = 0
x = 6 and x = -1/6
Speed can not be negative, so
x = 6 km/hr
Question 19
1 / -0

Joy buys some packets of colours for Rs. 300. If the cost of each packet had been decreased by Rs. 5, then he would have bought 5 more packets. How much does each packet cost?

A
Rs. 30

B
Rs. 25

C
Rs. 20

D
Rs. 15

Solution

Let the cost of each packet be Rs. x.
Number of packets bought for Rs. 300 = y

So, y = … (1)

If the cost of each packet is decreased by Rs. 5, then
Number of packets = y + 5 =

Comparing both the equations,

x² - 5x - 300 = 0
x = -15, 20
Negative value is rejected; so x = 20.
Question 20
1 / -0

A boat can travel a distance of 8.5 km in one hour in still water. If the same boat travels upstream for 72 km and then returns to the starting point, taking 17 hours for the entire journey, then what is the speed of the stream?

A
0.5 km/hr

B
0.25 km/hr

C
0.05 km/hr

D
0.025 km/hr

Solution

Let the speed of the stream be y (km/hr).
Using Speed = Distance/Time, we get:

1224 = (72.25 × 17) - 17y²
1228.25 - 1224 = 17y²
= y²
0.25 = y²
y = +0.5, -0.5
The speed of the stream cannot be negative, so the speed of the stream is 0.5 km/hr.
Question 21
1 / -0

How many positive values will 'x' take such that 4^{x + 1} - 18 × 2^x + 8 = 0?

A
0

B
1

C
2

D
More than 2

Solution

Assume 2^x = k.
The given equation will be 4k² - 18k + 8 = 0.
2k² - 9k + 4 = 0
k = k =
k = 4, So, if 2^x = 4,
x = 2
If 2^x = ,
x = -1
So, x will take only one positive value.
Question 22
1 / -0

Read the following statements carefully and choose the suitable.

Statement I: The quadratic equation ax² + bx + c = 0 has no real roots if b² - 4ac > 0.
Statement II: The number of possible values of m (m is a whole number) is 5 if the quadratic equation 4x² + 16x + 4m = 0 has real roots.

A
Both statements I and II are true.

B
Statement I is true, but statement II is false.

C
Statement I is false, but statement II is true.

D
Both statements I and II are false.

Solution

Statement I is false because if the quadratic equation ax² + bx + c = 0 has no real roots, then b² - 4ac < 0.
Statement II is true because for the equation 4x² + 16x + 4m = 0 to have real roots, b² – 4ac must be greater than or equal to 0, i.e. 256 - 64m ≥ 0 or 4 ≥ m, m being a whole number.
So, m can have 0, 1, 2, 3 and 4 as possible values. So, 5 values are possible for m.
Question 23
1 / -0

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and the rest of the park has been used as a lawn. What is the width of the road if the area of the lawn is 2109 m²?

A
2.91 m

B
3 m

C
6 m

D
None of these

Solution

Suppose, width of the road = x m

Area of the rectangular park = 60 × 40 = 2400 m²
Area of the two concrete crossroads
= 60 × x + 40 × x - x × x
= 60x + 40x - x²
= 100x - x²
Then,
2400 - (100x - x²) = 2109
2400 - 100x + x² = 2109
x² - 100x + 2400 - 2109 = 0
x² - 100x + 291 = 0
x² - 3x - 97x + 291 = 0
x(x - 3) - 97(x - 3) = 0
(x - 97)(x - 3) = 0
x = 97 (not valid) and x = 3 m

Question 24

1 / -0

State 'T' for true and 'F' for false for the following statements and choose the suitable option from the table given below them.

(i) If one of the roots of the quadratic equation 2x²- kx - 3 = 0 is -1, the other root is 5.
(ii) If α + β = 4 and α³ + β³ = 44, then α and β are the roots of 3x² - 12 x + 5 = 0.
(iii) If the roots of ax² + b = 0 are real and distinct, then ab > 0.
(iv) The quadratic equation (y - c)(y - d) + 2 = 0 has real roots if (c - d) ≥ (2)^3/2 and (c - d) ≤ -(2)^3/2.

	(i)	(ii)	(iii)	(iv)
(A)	F	T	F	F
(B)	F	T	F	T
(C)	F	T	T	F
(D)	F	T	T	T

(A)

(B)

(C)

(D)

Solution

(i) is false.
One root is -1;
So, 2(-1)² - k(-1) - 3 = 0
2 + k - 3 = 0
k = 1
Now, 2x² - x - 3 = 0
2x² - 3x + 2x - 3 = 0
(2x - 3)(x + 1) = 0
x = 3/2 and x = -1

(ii) is true.
α + β = 4
Or α³ + β³ = (α + β)³ - 3αβ(α + β) = 44
64 - 3αb (4) = 44
3ab(4) = 20
ab = 5/3 (Product of roots)
So, the equation is:
x² - 4x + 5/3 = 0
3x² - 12 x + 5 = 0

(iii) is false.
Since roots are real, discriminant (D) > 0
D > 0 or 0² - 4ab > 0
ab < 0

(iv) is true.
The given equation can be written as y² - (d + c)y + cd + 2 = 0.
For real roots, D = (d + c)² - 4(cd + 2) ≥ 0
Or d² + c² - 2cd - 8 ≥ 0
Or (c - d)² ≥ 8
Or (c - d) ≥ (2)^3/2 and (c - d) ≤ -(2)^3/2

Question 25
1 / -0

In the given figure, ABC is a right triangle with B as the right angle. What is the length of AC?

A
25 cm

B
26 cm

C
28 cm

D
30 cm

Solution

In ΔABC, (24)² + (x)² = (3x + 4)²
576 + x² = 9x² + 24x + 16
9x² + 24x + 16 – x² – 576 = 0
8x² + 24x – 560 = 0
Dividing by 8, we get
x² + 3x – 70 = 0
(x + 10) (x – 7) = 0
x = –10 not possible.
∴ x = 7
When x = 7,
AC = 3x + 4 = 3(7) + 4
∴ AC = 25 cm

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Quadratic Equations Test - 6

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Question 1

c = 0

c = 0, b ≠ 0

b = 0, c = 0

b ≠ 0, c ≠ 0

Solution

Question 2

-2 -

1/4

-2 +

Cannot be determined

Solution

Question 3

c >

c >

c >

None of these

Solution

Question 4

Solution

Question 5

infinitely many roots

two roots

one root

no root

Solution

Question 6

Solution

Question 7

28

17

11

7

Solution

Question 8

-1/2

1/3

-2

1

Solution

Question 9

a ≥ 1

a ≤ 1

a > -3 or a ≤ 1

a < -3 or a ≥ 1

Solution

Question 10

Solution

Question 11

2 : 1 : 3

3 : 2 : 1

3 : 1 : 2

1 : 2 : 3

Solution

Question 12

None of these

Solution

Question 13

bx2 + ax + b = 0

bx2 + ax – b = 0

bx2 – ax – b = 0

bx2 – ax + b = 0

Solution

Question 14

acx2 - bx - 1 = 0

acx2 – bx + 1 = 0

acx2 + bx + 1 = 0

acx2 + bx – 1 = 0

Solution

Question 15

Solution

Question 16

¹/₄

bx² + ax + b = 0

bx² + ax – b = 0

bx² – ax – b = 0

bx² – ax + b = 0

acx² - bx - 1 = 0

acx² – bx + 1 = 0

acx² + bx + 1 = 0

acx² + bx – 1 = 0

144 cm²

169 cm²

196 cm²

256 cm²