Theory of Equations Test

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Theory of Equations Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

What is the number of real roots of the equation A²/x + B²/(x – 1) = 1, where A and B are real numbers not equal to zero?

A
0

B
1

C
1 or 2

D
2

Solution

x cannot be 0 or 1 as the denominator cannot be 0.
Now, A²x - A²+ B²x = x²- x
Or x² - x - A²x - B²x + A² = 0
Or x² - (A² + B² + 1)x + A²= 0
Now, D = (A² + B² + 1)² - 4A²
=(A² + B² + 1)² - (2A)²
= (A² + B² + 1 - 2A)(A² + B² + 1 + 2A)
= ((A - 1)² + B²)((A + 1)² + B²)
As every term is in square, D > 0
Thus, the equation has real and distinct roots, hence two real roots. So, option (3) is correct.
Question 2
1 / -0

If the equations x² – kx – 21 = 0 and x² – 3kx + 35 = 0 have a common root, the value of k is

A
–4

B
4 or –4

C
5 or 4

D
6

Solution

Given equations:
x² – kx – 21 = 0 ... (1)
And x² – 3kx + 35 = 0 ... (2)
Equation (1) - equation (2) gives
x = 28/k as common root
Substituting the value of x in equation (1),

k² = 16
k = 4 or k = -4
Question 3
1 / -0

The number of real roots of = 9 is

A
0

B
2

C
1

D
4

Solution

= 3²
⇒ 2x² - 7x + 7 = 2
⇒ 2x² - 7x + 5 = 0
D = 9, (D > 0)
∴ The given equation has two distinct real roots.
Question 4
1 / -0

If one root of the equation x² + mx + 12 = 0 is 4 and the equation x² + mx + n = 0 has equal roots, then the value of n is

A

B

C
4

D

Solution

x² + mx + 12 = 0
x = 4 will satisfy this equation.
⇒ 16 + 4m + 12 = 0
⇒ m = –7
The other equation becomes x² – 7x + n = 0.
Its roots are equal.
So, b² = 4ac
⇒ 49 = 4n
Or n =
Question 5
1 / -0

If the roots of the equation ax² + bx + c = 0 are such that their sum is equal to the sum of their squares, then

A
2ac = ab + b²

B
2ac = a² + b²

C
2ac = ab + b³

D
None of these

Solution

Given α+ β = α² + β²
= (α + β)² - 2αβ
As α + β = - and αβ =
∴ -= - 2 ×
-ab = b² - 2ac
2ac = b² + ab
Hence, answer option (1) is correct.
Question 6
1 / -0

If the roots of the equation x² + a² = 8x + 6a are imaginary, then

A
-2 < a < 8

B
- < a < 2, 8 < a <

C
- < a < –2, 8 < a <

D
None of these

Solution

The roots of the equation are imaginary, so D < 0.
⇒ 64 – 4 × 1 × (a² – 6a) < 0
⇒ – 4a² + 24a + 64 < 0
⇒ a² – 6a – 16 > 0
⇒ a² – 8a + 2a – 16 > 0
⇒ (a – 8)(a + 2) > 0
⇒ - < a < –2, 8 < a <
Question 7
1 / -0

The values of x, which satisfy both the equations x² -1 ≤ 0 and x² -x - 2 ≥ 0 lie in

A
(- 1, 2)

B
(- 1, 1)

C
(1, 2)

D
{- 1}

Solution

(x + 1) (x - 1) ≤ 0 ⇒ - 1 ≤ x ≤ 1
(x - 2) (x + 1) ≥ 0 ⇒ x ≥ 2 and x ≤ - 1
Only x = - 1 satisfies both the equations.
Question 8
1 / -0

If and are the roots of the equation x² + x + 1 = 0, then the equation with roots ¹⁹ and ⁷ is

A
x²+ x + 1 = 0

B
x² - x + 1 = 0

C
x² + x - 1 = 0

D
x² - x - 1 = 0

Solution

= and = ²
Sum of roots = ¹⁹ + ⁷ = ¹⁹ + (²)⁷= + ² = -1
Product of roots = ¹⁹ × ⁷ = ¹⁹ × ¹⁴ = ³³ = 1 (∵³ = 1)
So, required equation is x² + x + 1 = 0
Question 9
1 / -0

The number of real solution of |x |² + 3| x | + 2 = 0 is

A
0

B
2

C
4

D
1

Solution

|x|² is always positive, 3|x| is also positive. As L.H.S. is always greater than zero.
Question 10
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 11
1 / -0

If a > 0, the roots of a(x² + 1) = x(a² + 1) are

A

B
a, a

C

D
None of these

Solution

ax²+ a = a²x + x
ax² - a²x - x + a = 0
ax(x - a) - 1(x - a) = 0
(x - a)(ax - 1) = 0
x = a and x =
Question 12
1 / -0

If the product of the roots of the equation mx² + 6x + (2m - 1) = 0 is -1, then the value of m is

A
1

B
-1

C

D

Solution

Product of roots = = -1
⇒ m =
Hence, answer option (3) is correct.
Question 13
1 / -0

If one root of the equation x² + px + q = 0 is double the other, then is

A
1

B

C

D

Solution

Sum = α + 2α= -p or 3α = -p … (1)
Product = 2α² = q ... (2)
From (1) and (2), we get 2 × = q
⇒
Hence, answer option (4) is correct.
Question 14
1 / -0

If the sum of the roots of the equation (m + 1)x² + 2mx + 3 = 0 is 1, then the value of m is

A

B

C

D

Solution

- = 1 m =
Hence, answer option (4) is correct.
Question 15
1 / -0

The equation x - = 1 - has

A
no root

B
one root

C
two roots

D
infinitely many roots

Solution

x - = 1 -
Solving we get x = 1 but x can`t be 1 (As denominator can`t be zero).
Question 16
1 / -0

The number of real roots of the equation (x – 1)² + (x – 2)² + (x – 3)² = 0 is

A
1

B
2

C
3

D
none of these

Solution

The equation has no real root because its LHS is always positive, while its RHS is zero.
Question 17
1 / -0

If and are the roots of the equation px² + qx + r = 0, then what is the value of ³ + ³?

A
(1/p³)(3pqr - q³)

B
(3pqr - q³)

C
p³(3pqr - q³)

D
3pqr

Solution

+ = -q/p and = r/p³ + ³ = ( + )(² + ² - )
= (- q/p)[( + )² - 3] = (1/p³)(3pqr - q³)
Hence, answer option (1) is correct.
Question 18
1 / -0

If one root of the equation ax² + bx + c = 0 is double the other, then

A
2b² = 9ac

B
b² = 4ac

C
b² = 9ac

D
None of these

Solution

Let one of the roots be .
So, the other will be 2.
Now, sum of roots = + 2 = -b/a 3 -b/a … (1)
And product of roots = 2² = c/a … (2)
From (1),
Substituting the value of in equation (2),
2b² = 9ac
Question 19
1 / -0

If , and are the roots of x³ - 2x² + x - 1 = 0, then the value of 1/ + 1/ + 1/ is

A
0

B
-1

C
1

D
4

Solution

+ + = ... (1)

If , and are the roots of x³ - 2x² + x - 1 = 0, then

From equation (1),
Question 20
1 / -0

For what value of m will the equation

(m + 1) x² + 2 (m + 3) x + m + 8 = 0 have equal roots?

A

B

C
-2

D

Solution

For equal roots D = 0 or b² - 4ac = 0
4 (m + 3)² - 4 (m + 1) (m + 8) = 0
Solving we get m =