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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!

Greg Martin
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emil
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3 Answers3

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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.

Alexander Burstein
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    This is a neat solution, if all coefficients of a quadratic polynom are real, then either both the roots are real or they are complex conjugate of each other – acat3 Apr 21 '20 at 05:33
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Hint: remarkably, that expression for $\Delta$ is a perfect square...! Can you factor the expression knowing that?

Greg Martin
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  • I'm trying to.. is there a specific method to select factors or is it just trial and error? Im trying $(a^2+....)(a^2+....)$ – emil Apr 21 '20 at 05:18
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    Note that $(a+b)(a-c)=(a^2-bc)+a(b-c)$, so your equation has the form $Ax^2+(A+C)x+C=0$ for some real $A$ and $C$. The left-hand side factors into $(x+1)(Ax+C)$. The discriminant, then, is $(A-C)^2=(a-b)^2(a+c)^2$. – Alexander Burstein Apr 21 '20 at 05:24
  • All of the answers comments helped thanks a lot – emil Apr 21 '20 at 05:47
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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.

Aman Gupta
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