Prove this has real roots $(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$

Question

Prove this has real roots

$(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$

My Work

\begin{align*} \Delta&=(a+b)^2(a-c)^2-4a(b-c)(a^2-bc) \\ &=a^4+2a^3c+a^2c^2-2a^3b+b^2a^2-4a^2bc-2abc^2+2ab^2c+b^2c^2. \end{align*}

How do I show that this is positive? Simplification doesn't help either...How to factor them? please help!

Answer 1

Note that $-1$ is a root, so the other root is also real. It turns out to be $\dfrac{a(b-c)}{bc-a^2}$, but that's not really necessary for what you need to prove.

Answer 2

Hint: remarkably, that expression for $\Delta$ is a perfect square...! Can you factor the expression knowing that?

Answer 3

If you put $x=-1$, in the equation you would see the LHS equals to $0$, it means $0$ is one of the root of the equation. Also, a quadratic equation have three possibilities of roots existence. 1. Real and distinct(both) 2. Real and equal 3. Imaginary roots(occurs in conjugate form, means if one is imaginary, other should be also an imaginary root)

In our case, we have obtained a real root, so other root must be real. Hence the given equation should have real roots, I think method of discriminant is a bit difficult. You can go by using one root method.