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.**If |z

^{2}-1|=|z|

^{2}+1, then z lies on

Solution

(d)

Let z = x+iy. Then,

|z

⇒ |(x

⇒ (x

(d)

Let z = x+iy. Then,

|z

^{2}-1|=|z|^{2}+1⇒ |(x

^{2}-y^{2}-1)+2ixy|=x^{2}+y^{2}+1⇒ (x

^{2}-y^{2}-1)^{2}+4x^{2}y^{2}=(x^{2}+y^{2}+1)^{2}⇒ x=0 Hence, z lies on imaginary axis

**Q4.**Let Î± , Î² be the roots of the equation x

^{2}-px+r=0 and Î±/2 , 2Î² be the roots of the equation x

^{2}-qx+r=0. Then the value of r is

Solution

(d)

Since, the equation x^2-px+r=0 has roots (Î±,Î²) and the equation x

∴ Î±+Î² = p and r = Î±Î² and Î±/2 + 2Î²=q

⇒ Î²=(2q-p)/3 and Î±=(2(2p-q))/3

∴ Î±Î²=r=2/9 (2q-p)(2p-q)

(d)

Since, the equation x^2-px+r=0 has roots (Î±,Î²) and the equation x

^{2}-qx+r=0 has roots (Î±/2,2Î²)∴ Î±+Î² = p and r = Î±Î² and Î±/2 + 2Î²=q

⇒ Î²=(2q-p)/3 and Î±=(2(2p-q))/3

∴ Î±Î²=r=2/9 (2q-p)(2p-q)

**Q5.**For x

^{2}-(a+3) |x|+4=0 to have real solutions, the range of a is

Solution

(d)

a = (x

= |x| + 4/|x| - 3 = (√(|x| ) - 2/√(|x| ))+1

⇒ a ≥ 1

(d)

a = (x

^{2}+4)/(|x|)-3= |x| + 4/|x| - 3 = (√(|x| ) - 2/√(|x| ))+1

⇒ a ≥ 1

**Q7.**The number of real solutions of the equation (9/10)

^{x}=-3+x-x

^{2}is

Solution

(b) Let f(x)=-3+x-x

(b) Let f(x)=-3+x-x

^{2}. Then f(x) < 0 for all x, because coefficient of x^{2}is less than 0 and D< 0. Thus, L.H.S. of the given equation is always positive whereas the R.H.S. is always less than zero. Hence, there is no solution**Q8.**If p,q,r are +ve and are in A.P., in the roots of quadratic equation px

^{2}+qx+r=0 are all real for

Solution

(b) For real roots, q

(b) For real roots, q

^{2}-4pr≥0 ⇒((p+r)/2)^{2}-4pr≥0 (∵p,q,r are in A.P.) ⇒p^{2}+r^{2}-14pr≥0 ⇒p^{2}/r^{2}-14 p/r+1≥0 ⇒(p/r-7)^{2}-48≥0 ⇒|p/r-7|≥4√3**Q9.**If arg(z)< 0 , then arg(-z) - arg(z)=

Solution

(a)

arg(-z)- arg(z) = arg((-z)/z) = arg(-1)=Ï€

(a)

arg(-z)- arg(z) = arg((-z)/z) = arg(-1)=Ï€

**Q10.**If Î± , Î² , Î³ , Ïƒ are the roots of the equation x

^{4}+4x

^{3}-6x

^{2}+7x-9=0, then the value of (1+Î±

^{2})(1+Î²

^{2})(1+Î³

^{2})(1+Ïƒ

^{2}) is