Question

Let $\alpha, \beta$ be the roots of the equation $x^2-\sqrt{2} x+\sqrt{6}=0$ and $\frac{1}{\alpha^2}+1, \frac{1}{\beta^2}+1$ be the roots of the equation $x^2+a x+b=0$. Then the roots of the equation $x^2-(a+b-2)$ $x +( a + b +2)=0$ are :

• A. non-real complex numbers
• B. real and both negative
• C. real and both positive
• D. real and exactly one of them is positive

Answer 1

Answer: (B)

$ a=\frac{-1}{\alpha^2}-\frac{1}{\beta^2}-2$
$ b=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+1+\frac{1}{\alpha^2 \beta^2}$
$ a+b=\frac{1}{(\alpha \beta)^2}-1=\frac{1}{6}-1=-\frac{5}{6} $
$x^2-\left(-\frac{5}{6}-2\right) x+\left(2-\frac{5}{6}\right)=0 $
$ 6 x^2+17 x+7=0 $
$ x=-\frac{7}{3}, x=-\frac{1}{2} $ are the roots
Both roots are real and negative.