Question

Let p,q be the roots of the equations x^2-3ax+a^2=0. Determine a if we are given that p^2+q^2=7/4

Answer 1

By using the relation

$p+q=3a$

$pq=a^2$

According to the identity

$(p+q)^2=p^2+q^2+2pq$

Now subtract $2pq$

$(p+q)^2-2pq=p^2+q^2$

By using the identity above

$p^2+q^2=(3a)^2-2a^2=7a^2$

Given that

$7a^2=\frac{7}{4}$

Hence

$a=\pm\frac{1}{2}$