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)..

**Q3.**If Î±,Î²,Î³ are the roots of x^3 -x^2 -1=0 then the value of (1+Î±)/(1-Î±)+(1+Î²)/(1-Î²)+(1+Î³)/(1-Î³) is equal to

Solution

(a) ∑ Î±=1,∑ Î±Î²=0, Î±Î²Î³ = 1

∑(1+Î±) / (1-Î±) = -∑(-Î±+1-2)/(1-Î±) = ∑(2/(1-Î±)-1)

= 2∑〖1/(1-Î±)-3〗

Now,

1/((x-Î±)) +1/((x-Î²)) +1/((1-Î³)) = ( 3x^2- 2x)/( x^3 -x^2 -1)

⇒ 1/(1-Î±) +1/(1-Î²) +1/(1-Î³) =(3-2)/(1-1-1)=-1

⇒ (1+Î±)/(1-Î±)=-5

(a) ∑ Î±=1,∑ Î±Î²=0, Î±Î²Î³ = 1

∑(1+Î±) / (1-Î±) = -∑(-Î±+1-2)/(1-Î±) = ∑(2/(1-Î±)-1)

= 2∑〖1/(1-Î±)-3〗

Now,

1/((x-Î±)) +1/((x-Î²)) +1/((1-Î³)) = ( 3x^2- 2x)/( x^3 -x^2 -1)

⇒ 1/(1-Î±) +1/(1-Î²) +1/(1-Î³) =(3-2)/(1-1-1)=-1

⇒ (1+Î±)/(1-Î±)=-5

**Q4.**If the equation |x^2+bx+c|=k has four roots, then

**Q6.**If a(p+q)^2+2bpq+c=0 and a(p+r)^2+2bpr+c=0 (a≠0), then

Solution

(b) Given,

a(p+q)^2+2bpq+c=0 and a(p+r)^2+2bpr+c=0

⇒q and r satisfy the equation a(p+x)^2+2bpx+c=0

⇒q and r are the roots of ax^2 + 2(ap +bp )x + c+ ap^2 = 0

⇒qr= product of roots =(c+ ap^2)/a = p^2 + c/a

(b) Given,

a(p+q)^2+2bpq+c=0 and a(p+r)^2+2bpr+c=0

⇒q and r satisfy the equation a(p+x)^2+2bpx+c=0

⇒q and r are the roots of ax^2 + 2(ap +bp )x + c+ ap^2 = 0

⇒qr= product of roots =(c+ ap^2)/a = p^2 + c/a

**Q7.**The value of m for which one of the roots of x^2 -3x+ 2m = 0 is double of one of the roots of x^2-x+m=0 is

Solution

(a) Let Î± be the root of x^2-x+m=0 and 2Î± be the root of x^2-3x+2m=0.

Then, Î±^2-Î±+m=0 and 4Î±^2-6Î±+2m=0

Eliminating Î±,m^2=-2m

⇒m=0,m=-2

(a) Let Î± be the root of x^2-x+m=0 and 2Î± be the root of x^2-3x+2m=0.

Then, Î±^2-Î±+m=0 and 4Î±^2-6Î±+2m=0

Eliminating Î±,m^2=-2m

⇒m=0,m=-2

**Q9.**Total number of integral values of ‘a’ so that x^2-(a+1)x+a-1=0 has integral roots is equal to

Solution

(a) x^2-(a+1)x+a-1=0

⇒(x-a)(x-1)=1

Now, a∈I and we want x to be an integer.

Hence, x-a=1,x-1=1 or x-a=-1,x-1=-1

⇒a=1 in both cases

(a) x^2-(a+1)x+a-1=0

⇒(x-a)(x-1)=1

Now, a∈I and we want x to be an integer.

Hence, x-a=1,x-1=1 or x-a=-1,x-1=-1

⇒a=1 in both cases

**Q10.**z_1,z_2,z_3,z_4 are distinct complex numbers representing the vertices of a quadrilateral ABCD taken in order. If z_1-z_4=z_2-z_3 and arg〖[(z_4-z_1)/(z_2-z_1)]〗=Ï€/2, then the quadrilateral is: