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 $x^2+ax+b=0$. 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)

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