If $\alpha, \beta$ are roots of $x^2-3ax+a^2=0$, find the value(s) of $a$ if $\alpha^2+\beta^2=\frac{7}{4}$.

Question

If $\alpha, \beta$ are roots of $x^2-3ax+a^2=0$, how do you find the value(s) of a if $\alpha^2+\beta^2=\frac{7}{4}$.

• I tried to substitute the $\alpha, \beta$ but confused since it is equal to $0$.
• Tried to expand $\alpha^2+\beta^2$. No clue here too.

How do you solve this?

Answer 1

Viète's relations give us $\alpha+\beta=3a$ and $\alpha\beta=a^2$, from which we get an equation in $a$: $$(3a)^2-2a^2=(\alpha+\beta)^2-2\alpha\beta=\alpha^2+\beta^2=\frac74$$ This simplifies to $7a^2=\frac74$ or $a^2=\frac14$, whence $a=\pm\frac12$.

Answer 2

We know that: $\alpha+\beta=3a$ and $\alpha\beta=a^2$. Now, $$(\alpha+\beta)^2-2\alpha\beta=\alpha^2+\beta^2$$ So, $(\alpha+\beta)^2-2\alpha\beta=(3a)^2-2a^2=\frac74$

Now, just equate:

$7a^2=\frac74 \implies a^2=\frac14$, then $a=\pm\frac12$.

Answer 3

No squaring: $${7\over 4}=\alpha^2+\beta^2 =(3a\alpha^2-a^2)+(3\beta -a^2) = 3a\underbrace{(\alpha+\beta)}_{3a}-2a^2 =7a^2$$

So $a=\pm{1\over 2}$.