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Mathematics Question on Complex Numbers and Quadratic Equations

If 1α\frac{1}{\sqrt{\alpha}} and 1β\frac{1}{\sqrt{\beta}} are the roots of the equation, ax2+bx+1=0(a≠0,a,b∈R)ax^{2} + bx +1 = 0 \left(a ^{ }\ne 0, a, b \in R\right), then the equation, x(x+b3)+(a3?3abx)x\left(x + b^{3}\right) + \left(a^{3} ? 3abx\right) = 0 has roots :

A

α3/2\alpha^{3/2} and β3/2\beta^{3/2}

B

α β1/2\alpha \,\beta^{1/2} and α1/2 β\alpha^{1/2}\,\beta

C

α β\sqrt{\alpha \,\beta } and α β\alpha\,\beta

D

α−32\alpha^{-\frac{3}{2}} and β−32\beta^{-\frac{3}{2}}

Answer

α3/2\alpha^{3/2} and β3/2\beta^{3/2}

Explanation

Solution

1α+1β=−baalso1αβ=1a\frac{1}{\sqrt{\alpha}}+\frac{1}{\sqrt{\beta}}=-\frac{b}{a} also \frac{1}{\sqrt{\alpha\beta }}=\frac{1}{a}
⇒α+β=−b\Rightarrow \sqrt{\alpha}+\sqrt{\beta}=-b
now x(x+b3)+a3−3abxx \left(x + b^{3}\right) + a^{3} - 3abx
=x2+(b3−3ab)x+a3=x^{2} + \left(b^{3} - 3ab\right) x + a^{3}
=x2+b(b2−3a)x+a3= x^{2}+ b \left(b^{2} - 3a\right) x +a^{3}
=x^{2}-\left(\sqrt{\alpha }+\sqrt{\beta }\right)\left\\{\alpha+\beta+2\sqrt{\alpha\beta }-3\sqrt{\alpha \beta }\right\\}x+\alpha\beta \sqrt{\alpha\beta }
=x2−(αα+ββ)+αβαβ=x^{2}-\left(\alpha\sqrt{\alpha }+\beta\sqrt{ \beta }\right)+\alpha\beta\sqrt{\alpha \beta }
⇒\Rightarrow roots are αα\alpha \sqrt{\alpha} and ββ\beta \sqrt{ \beta}