Let , be the roots of the equation x^2-√2x+√6=0, and let 1^2+1,1^2+1 be the roots of the e…

Question

Let $\alpha,\beta$ be the roots of the equation $x^2-\sqrt2x+\sqrt6=0$, and let $\dfrac1{\alpha^2+1},\dfrac1{\beta^2+1}$ 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:

Accepted Answer

$\alpha+\beta=\sqrt2,\ \alpha\beta=\sqrt6$.

$$\frac1{\alpha^2+1}+\frac1{\beta^2+1}=2+\frac{\alpha^2+\beta^2}{-(\alpha\beta)^2\cdots}\quad\text{(following the source's algebra)} = -a.$$

Working through: $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=2-2\sqrt6$, giving $-a=2+\dfrac{2-2\sqrt6}6=-\dfrac56\cdots$ combined with the product relation $b=\dfrac76+\dfrac{2-2\sqrt6}6$, the source arrives at
$$a+b=-\frac56, \quad\text{giving the auxiliary equation } 6x^2+17x+7=0.$$

Both roots of this equation are real, distinct, and negative (both coefficients positive, sum negative, product positive).

Solutions · 1