Complex numbers and quadratic equations is a segment of maths that deals with crucial theorems and concepts along with various formulae. It comprises of linear and quadratic equations along with roots related to the complex number's set (known as complex roots)..

**Q1.**Let f(x) = ax^2 - bx+ c^2 , b ≠ 0 and f(x) ≠ 0 for all x ∈ R. Then:

Solution

(b)

Here, ax^2 - bx + c^2 = 0 does not have real roots. So,

D < 0 ⇒ b^2 - 4ac^2 < 0 ⇒ a > 0

Therefore, f(x) is always positive. So,

f(2) > 0 ⇒ 4a - 2b + c^2 > 0

(b)

Here, ax^2 - bx + c^2 = 0 does not have real roots. So,

D < 0 ⇒ b^2 - 4ac^2 < 0 ⇒ a > 0

Therefore, f(x) is always positive. So,

f(2) > 0 ⇒ 4a - 2b + c^2 > 0

**Q2.**The number of real roots of the equation x^2 - 3| x | + 2 = 0 is

Solution

(c)

We have,

|x|^2 - 3|x| + 2 = 0

⇒(|x| - 1)(|x| - 2) = 0

⇒|x| = 1 , 2 ⇒ x = ±1 , ±2

(c)

We have,

|x|^2 - 3|x| + 2 = 0

⇒(|x| - 1)(|x| - 2) = 0

⇒|x| = 1 , 2 ⇒ x = ±1 , ±2

**Q3.**If z_1 and z_2 are the complex roots of the equation (x - 3)^3 + 1 = 0, then z_1 + z_2 equals to

Solution

(d)

(x - 3)^3 + 1 = 0

⇒((x - 3)/ -1))^3 = 1

⇒(x - 3)/( -1 ) = 1 , Ï‰ , Ï‰^2

⇒x=2 , 3 - Ï‰ , 3 - Ï‰^2

Hence, the run of complex root is 6 - (Ï‰ + Ï‰)^2 = 6 + 1 = 7

(d)

(x - 3)^3 + 1 = 0

⇒((x - 3)/ -1))^3 = 1

⇒(x - 3)/( -1 ) = 1 , Ï‰ , Ï‰^2

⇒x=2 , 3 - Ï‰ , 3 - Ï‰^2

Hence, the run of complex root is 6 - (Ï‰ + Ï‰)^2 = 6 + 1 = 7

**Q4.**If x^2 + px + 1 is factor of the expression ax^3 + bx + c , then

Solution

(a)

Given that x^2 + px + 1 is a factor of ax^3 + bx + c. Then let ax^3 + bx + c = (x^2 + px + 1)(ax + Î»), where Î» is a constant. Then equation the coefficients of like powers of x on both sides, we get

0 = ap + Î» , b = pÎ» + a ,c = Î»

⇒p = -Î»/a = -c/a

Hence,

b = (-c/a)c + a

or ab = a^2 - c^2

(a)

Given that x^2 + px + 1 is a factor of ax^3 + bx + c. Then let ax^3 + bx + c = (x^2 + px + 1)(ax + Î»), where Î» is a constant. Then equation the coefficients of like powers of x on both sides, we get

0 = ap + Î» , b = pÎ» + a ,c = Î»

⇒p = -Î»/a = -c/a

Hence,

b = (-c/a)c + a

or ab = a^2 - c^2

Solution

(a)

i = cos[Ï€/2] + i sin[Ï€/2] = e^(iÏ€/2)

⇒i^i = ( e^(iÏ€/2) )^i = e^(-Ï€/2)

⇒z= (i)^((i)^i ) = [i^e]^(-Ï€/2)

⇒|z| = 1

(a)

i = cos[Ï€/2] + i sin[Ï€/2] = e^(iÏ€/2)

⇒i^i = ( e^(iÏ€/2) )^i = e^(-Ï€/2)

⇒z= (i)^((i)^i ) = [i^e]^(-Ï€/2)

⇒|z| = 1

**Q6.**The number of roots of the equation √(x - 2) (x^2 - 4x + 3 ) = 0 is

Solution

(d) Given equation is satisfied by x = 1 , 2 , 3. But for x = 1 , √(x - 2) is not defined. Hence, number of roots is 2 and the roots are x=2 and 3

(d) Given equation is satisfied by x = 1 , 2 , 3. But for x = 1 , √(x - 2) is not defined. Hence, number of roots is 2 and the roots are x=2 and 3

**Q7.**Total number of values of a so that x^2 - x - a = 0 has integral roots, where a ∈ N and 6 ≤ a ≤ 100, is equal to

Solution

(d)

x^2 - x - a = 0 ,D = 1 + 4a = odd

D must be perfect square of some odd integer. Let

D = (2Î» + 1)^2

⇒ 1 + 4a = 1 + 4Î»^2 + 4Î»

⇒ a = Î»(Î» + 1)

Now,

a ∈ [6 , 100]

⇒ a = 6 , 12 , 20 , 30 , 42 , 56 , 72 , 90

Thus a can attain eight different values

(d)

x^2 - x - a = 0 ,D = 1 + 4a = odd

D must be perfect square of some odd integer. Let

D = (2Î» + 1)^2

⇒ 1 + 4a = 1 + 4Î»^2 + 4Î»

⇒ a = Î»(Î» + 1)

Now,

a ∈ [6 , 100]

⇒ a = 6 , 12 , 20 , 30 , 42 , 56 , 72 , 90

Thus a can attain eight different values

**Q8.**If a,b,care the sides of the triangle ABC such that a ≠ b ≠ c and x^2 - 2(a + b + c)x + 3Î»(ab + bc+ ca ) = 0 has real roots, then

Solution

(a)

Since, a,b and c are the sides of a ∆ ABC, then

|a - b| < |c| ⇒ a^2 + b^2 - 2ab < c^2

Similarly, b^2 + c^2 - 2bc < a^2, c^2 + a^2 - 2ca < b^2

On adding, we get

(a^2 + b^2 + c^2 ) < 2(ab + bc + ca)

⇒ (a^2 + b^2 + c^2) / (ab + bc + ca )< 2 …..(i)

Also, D ≥ 0 ,(a + b + c)^2 - 3Î»(ab + bc + ca) ≥ 0

⇒ (a^2 + b^2 + c^2) / ( ab + bc + ca) > 3Î»-2 ………(ii)

From Eqs. (i) and (ii),

3Î» - 2 < 2 ⇒ Î» < 4/3

(a)

Since, a,b and c are the sides of a ∆ ABC, then

|a - b| < |c| ⇒ a^2 + b^2 - 2ab < c^2

Similarly, b^2 + c^2 - 2bc < a^2, c^2 + a^2 - 2ca < b^2

On adding, we get

(a^2 + b^2 + c^2 ) < 2(ab + bc + ca)

⇒ (a^2 + b^2 + c^2) / (ab + bc + ca )< 2 …..(i)

Also, D ≥ 0 ,(a + b + c)^2 - 3Î»(ab + bc + ca) ≥ 0

⇒ (a^2 + b^2 + c^2) / ( ab + bc + ca) > 3Î»-2 ………(ii)

From Eqs. (i) and (ii),

3Î» - 2 < 2 ⇒ Î» < 4/3

**Q9.**Suppose A,B,C are defined as A=a^2 b+ab^2-a^2 c-ac^2,B=b^2 c+bc^2-a^2 b-ab^2 and C=a^2 c+ac^2-b^2 c-bc^2, where a>b>c>0 and the equation Ax^2+Bx+C=0 has equal roots, then a,b,c are in

Solution

(c)

A = a(b - c)(a + b + c)

B = b(c - a)(a + b + c)

C = c(a - b)(a + b + c)

Now,

Ax^2 + Bx + C = 0

⇒(a + b + c) {a(b - c) x^2 + b(c - a)x + c(a - b)} = 0

Given that roots are equal. Hence,

D = 0

⇒b^2 (c - a^2 ) - 4ac(b - c)(a - b) = 0

⇒b^2 c^2 - 2ab^2 c + b^2 a^2 - 4a^2 bc + 4acb^2 + 4a^2 c^2- 4abc^2 = 0

⇒b^2 c^2 + b^2 a^2 + 4a^2 c^2 + 2ab^2 c - 4a^2 bc - 4abc^2 = 0

⇒(bc + ab - 2ac)^2 = 0

⇒bc + ab = 2ac

⇒1/a + 1/c = 2/b

⇒a , b , c are in H.P.

(c)

A = a(b - c)(a + b + c)

B = b(c - a)(a + b + c)

C = c(a - b)(a + b + c)

Now,

Ax^2 + Bx + C = 0

⇒(a + b + c) {a(b - c) x^2 + b(c - a)x + c(a - b)} = 0

Given that roots are equal. Hence,

D = 0

⇒b^2 (c - a^2 ) - 4ac(b - c)(a - b) = 0

⇒b^2 c^2 - 2ab^2 c + b^2 a^2 - 4a^2 bc + 4acb^2 + 4a^2 c^2- 4abc^2 = 0

⇒b^2 c^2 + b^2 a^2 + 4a^2 c^2 + 2ab^2 c - 4a^2 bc - 4abc^2 = 0

⇒(bc + ab - 2ac)^2 = 0

⇒bc + ab = 2ac

⇒1/a + 1/c = 2/b

⇒a , b , c are in H.P.

**Q10.**Consider the equation x^2 + 2x - n = 0, where n ∈ N and n ∈ [15 , 100]. Total number of different values of ‘n’ so that the given equation has integral roots is

Solution

(a)

x^2 + 2x - n = 0

⇒(x + 1)^2 = n + 1

⇒x = -1 ± √(n + 1)

Thus, n+1 should be a perfect square. Now,

n ∈ [5 , 100]

⇒ n + 1 ∈[6 , 101]

Perfect square values of n+1 are 9 ,16 ,25 ,36 ,49 ,64 ,81 ,100. Hence, number of values is 8

(a)

x^2 + 2x - n = 0

⇒(x + 1)^2 = n + 1

⇒x = -1 ± √(n + 1)

Thus, n+1 should be a perfect square. Now,

n ∈ [5 , 100]

⇒ n + 1 ∈[6 , 101]

Perfect square values of n+1 are 9 ,16 ,25 ,36 ,49 ,64 ,81 ,100. Hence, number of values is 8