THERORY OF EQAUTIONS PRACTICE PROBLEMS

Problem 1 :

A zero of x³ + 64 is

(1) 0     (2)  4    (3)  4i     (4)  -4

Solution :

x³ + 64 = 0

x³ = -64

x³ = (-4)³

Powers are equal, then equate the bases.

x = -4

So, option (4) is correct.

Problem 2 :

If f and g are polynomials of degree m and n respectively and if h(x) = (f o g)(x), then the degree of h is

(1) mn     (2)  m+n   (3)  mⁿ     (4) n^m

Solution :

Let f(x) = x + 2

Degree = 1

g(x) = x² + 3

degree = 2

(fog)(x) = f[g(x)]

= f[x² + 3]

= x² + 3 + 2

degree of (fog)(x) = degree of f(x) x degree of g(x)

= 2 x 1

= 2

Degree of (fog)(x) is mn.

Problem 3 :

A polynomial equation in x of degree n always has 

(1)  n distinct roots      (2) n real roots    (3)  n complex roots

(4) at most one root

Solution :

The polynomial which has the highest exponent of n will have n roots. Those roots may be

• sometimes real.
• sometimes imaginary
• sometimes real and imaginary

In this case options (2), (3) and (4) are incorrect. So, option (1) is correct.

Problem 4 :

If α, β and γ are the zeros of x³ + px² + qx + r, then Σ 1/α is

(1)  -q/r      (2) -p/r    (3)  q/r     (4) -q/r

Solution :

Σ 1/α = 1/α + 1/β + 1/γ

= (βγ + αγ + βα) / αβγ

= (αβ + βγ + αγ) / αβγ -----(1)

x³ + px² + qx + r

Comparing the given equation with  ax³ + bx² + cx + d

a = 1, b = p, c = q and d = r

αβ + βγ + αγ = c/a

= q/1

αβ + βγ + αγ = q

αβγ = -d/a

= -r/1

αβγ = -r

Applying the above values in (1), we get

= q / (-r)

= -q/r

So, option (1) is correct.

Problem 5 :

According to the rational root theorem, which number is not possible rational zero of

4x⁷ + 2x⁴ - 10x³ - 5 ?

(1)  -1      (2) 5/4    (3)  4/5     (4) 5

Solution :

According to our notations, a_n = 4 and a₀ = -5. If p/q is a zero of the polynomial, then as (p, q) = 1, p must divide 5 and q must divide 4. 

Possible values of p are ±1 and ±5

Possible values of q are ±1, ±2 and ±4

Possible roots q are ±1, ±5, ±1/2, ±5/2, ±1/4, ±5/4

In the above roots, 4/5 is not one of the values. So, option (3) is correct.

Problem 6 :

The polynomial x³ - kx² + 9x has three real zeros if and only if k satisfies

(1)  |k| ≤ 6        (2) |k| = 0    (3)  |k| > 6     (4) |k|  ≥ 6

Solution :

Let P(x) = x³ - kx² + 9x

= x(x² - kx + 9)

b² - 4ac = (-k)² - 4(1)(9)

= k² - 36

Using nature of roots,

• b² - 4ac > 0, then the roots are real.
• b² - 4ac = 0, then the roots are equal.
• b² - 4ac < 0, then the roots are unreal.

b² - 4ac ≥ 0

k² - 36 ≥ 0

k² ≥ 36

k ≥ 6

So, option (4) is correct.

Problem 7 :

The number of real numbers in [0, 2π] satisfying

sin⁴x - 2sin²x + 1 is

(1)  2       (2) 4    (3)  1     (4) ∞

Solution :

sin⁴x - 2sin²x + 1 = 0

Let t = sin²x

(sin²x)² - 2sin²x + 1 = 0

t² - 2t + 1 = 0

(t - 1)(t - 1) = 0

t = 1 and t = 1

 sin²x = 1

sin x = √1

sin x = ±1

x = sin^-1(1) and x = sin^-1(-1)

x = π/2 and x = 3π/2

The above equation has two solutions.

So, the answer is 2, option (1).

Problem 8 :

If x³ + 12x² + 10ax + 1999 definitely has a positive zero, if and only if

(1)  a ≥ 0        (2) a > 0    (3)  a < 0     (4) a ≤ 0

Solution :

Let p(x) = x³ + 12x² + 10ax + 1999

The above cubic polynomial will have 3 zeros.

Using descartes rule,

• if a > 0, then there is no changes in sign. Then 0 positive roots will be there.
• if a < 0, then the number of sign changes will be 2. So, it will have 2 real roots.

So, the answer is a < 0 , option (3).

Problem 9 :

The polynomial x³ + 2x + 3 has

(1) one negative and two imaginary zeros

(2)  one positive and two imaginary zeros

(3) three real zeros     (4)  no zeros.

Solution :

Let p(x) = x³ + 2x + 3

No sign changes, there is no positive zeros.

p(-x) = (-x)³ + 2(-x) + 3

p(-x) = -x³ - 2x + 3

Number of sign changes = 1

There is 1 negative root.

0 positive roots, 1 negative roots and 2 imaginary roots.

So, option (1) is correct.

Problem 10 :

The number of positive zeros of the polynomial

is

(1)  0        (2) n    (3)  < n     (4) r

Solution :

Since the highest exponent of the polynomial is n, it will have n zeros.