Let a,b R be such that the equation ax^2-2bx+15=0 has a repeated root .

Question

Let $a,b\in\mathbb R$ be such that the equation $ax^2-2bx+15=0$ has a repeated root $\alpha$. If $\alpha$ and $\beta$ are the roots of the equation $x^2-2bx+21=0$, then $\alpha^2+\beta^2$ is equal to:

Accepted Answer

Repeated root of $ax^2-2bx+15=0$: discriminant $=0 \implies 4b^2=60a \implies b^2=15a$, and $\alpha=\dfrac ba$ (repeated root $=\dfrac{2b}{2a}$).

Since $a=\dfrac{b^2}{15}$: $\alpha=\dfrac{b}{b^2/15}=\dfrac{15}b$, so $\alpha^2=\dfrac{225}{b^2}$.

Since $\alpha$ is also a root of $x^2-2bx+21=0$: $\alpha^2=2b\alpha-21=2b\left(\dfrac{15}b\right)-21=30-21=9$.

$$\frac{225}{b^2}=9 \implies b^2=25.$$

For $x^2-2bx+21=0$: $\alpha+\beta=2b,\ \alpha\beta=21$.

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=4b^2-42=4(25)-42=58.$$

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