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  4. If the roots of the equation a x2+x+b=0 be real and unequal where a, b ∈ R, then the roots of the equation x2-4 √a b x+1=0 will be

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Q. If the roots of the equation $a x^2+x+b=0$ be real and unequal where $a, b \in R$, then the roots of the equation $x^2-4 \sqrt{a b} x+1=0$ will be

Complex Numbers and Quadratic Equations

Solution:

$a x^2+x+b=0$
roots are real and unequal so $1-4 a b>0$ .......(i)
$ x^2-4 \sqrt{a b} x+1=0 $
$ D=16 a b-4=4[4 a b-1]<0 \text { \{from (i) }\}$
So roots are imaginary