Discriminant question

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.

Answer 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 $\lambda=\frac34$.

Answer 2

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

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

Answer 3

We can also arrive at this result without referring to the discriminant. With the terms "collected" as $ \ (4+3\lambda)·a^2 \ - \ 10 ·(1+\lambda)·ab \ + \ (10+3\lambda)·b^2 \ \ , $ we consider what is required in order for this expression to be a binomial-square of the form $ \ (\zeta·a \ - \ \eta·b)^2 \ \ . $ This leads us to $$ \zeta^2 \ \ = \ \ 4 \ + \ 3·\lambda \ \ , \ \ 2·\zeta·\eta \ \ = \ \ 10 ·(1 \ + \ \lambda) \ \ , \ \ \eta^2 \ \ = \ \ 10 \ + \ 3·\lambda \ \ . $$ We can avoid the discriminant explicitly by writing $$ \frac{ \zeta·\eta }{\zeta^2} \ \ = \ \ \frac{\eta}{\zeta} \ \ = \ \ \frac{ 5 ·(1 \ + \ \lambda) }{ 4 \ + \ 3·\lambda} \ \ , \ \ \frac{ \zeta·\eta }{\eta^2} \ \ = \ \ \frac{\zeta}{\eta} \ \ = \ \ \frac{ 5 ·(1 \ + \ \lambda) }{ 10 \ + \ 3·\lambda} \ \ . $$ Then $$ \frac{\eta}{\zeta} · \frac{\zeta}{\eta} \ \ = \ \ 1 \ \ = \ \ \frac{ 5 ·(1 \ + \ \lambda) }{ 4 \ + \ 3·\lambda} \ · \ \frac{ 5 ·(1 \ + \ \lambda) }{ 10 \ + \ 3·\lambda} $$ $$ \Rightarrow \ \ 25 ·(1 \ + \ \lambda)^2 \ \ = \ \ (4 \ + \ 3·\lambda) · (10 \ + \ 3·\lambda) $$ [Naturally, it turns up implicitly in the end...] $$ \Rightarrow \ \ 16·\lambda^2 \ + \ 8·\lambda \ - \ 15 \ \ = \ \ (4·\lambda \ + \ 5 )·(4·\lambda \ - \ 3) \ \ = \ \ 0 \ \ . $$

The two values for $ \ \lambda \ $ produce the polynomials $$ \mathbf{\lambda \ = \ -\frac54 \ \ :} \quad \quad \frac14·a^2 \ + \ \frac52·ab \ + \ \frac{25}{4}·b^2 \ \ = \ \ \left( \ \frac{a}{2} \ + \ \frac{5·b}{2} \ \right)^2 \ \ , $$ $$ \mathbf{\lambda \ = \ \frac34 \ \ :} \quad \quad \frac{25}{4}·a^2 \ - \ \frac{35}{2}·ab \ + \ \frac{49}{4}·b^2 \ \ = \ \ \left( \ \frac{5·a}{2} \ - \ \frac{7·b}{2} \ \right)^2 \ \ . $$