Let a,b R, a0, be such that ax^2-2bx+5=0 has a repeated root , which is also a root of x^2…

Question

Let $a,b\in\mathbb R$, $a
e0$, be such that $ax^2-2bx+5=0$ has a repeated root $\alpha$, which is also a root of $x^2-2bx-10=0$. If $\beta$ is the other root of this equation, then $\alpha^2+\beta^2$ is equal to:

Accepted Answer

Repeated root: $\alpha=\dfrac ba$, and $\alpha^2=\dfrac5a \implies b^2=5a$.

Since $\alpha$ also satisfies $x^2-2bx-10=0$: $b^2-2ab^2-10a^2=0$. Substituting $b^2=5a$: $5a-10a^2-10a^2=0 \implies 20a^2=5a \implies a=\dfrac14$, so $b^2=\dfrac54$.

For $x^2-2bx-10=0$: $\alpha^2\beta^2=-10\cdot$..., working through: $\alpha^2=20,\beta^2=5$.

$$\alpha^2+\beta^2=20+5=25.$$

Solutions · 1