Discriminant question

Find the values of $\lambda$ for which $4a^2-10ab+10b^2+\lambda(3a^2-10ab+3b^2)$ is a perfect square.

What i've tried: Rearrange to make it an expression about $a$

$(4+3\lambda)a^2-10b(1+\lambda)a+b^2(10+3\lambda)$

Then do $D=0$

$D/4=[5b(1+\lambda)]^2-b^2(4+3\lambda)(10+3\lambda)=0$

That gives you an equation with $\lambda^2$ and $b^2$, and if do the same by treating the original expression as an expression about $b$ you would simply get another equation with $\lambda^2$ and $a^2$,and I don't see how you can get the values for $\lambda$ from there.

The textbook has the answer $\lambda=-5/4$ and $3/4$

Thanks.

Solution 1:

What you did is fine. Now, note that$$\frac D4=b^2\bigl(25(1+\lambda)^2-(4+3\lambda)(10+3\lambda)\bigr).$$So, solve the quadratic equation$$25(1+\lambda)^2=(4+3\lambda)(10+3\lambda);$$you will get that $\lambda=-\frac54$ or that $\lambda=\frac34$.

Solution 2:

Taking your expression about $a$, it's determinant must be $0$.

So we have $D=(10b(1+\lambda))^{2}-4b^{2}(4+3\lambda)(10+3\lambda)=b^{2}(100(1+\lambda)^{2}-4(4+3\lambda)(10+3\lambda))=b^{2}(64\lambda^{2}+32\lambda-60)=4b^{2}(16\lambda^{2}+8\lambda-15)=4b^{2}(4\lambda-3)(4\lambda+5)=0.$