High school level question of two quadratic equations sharing a common root.

Question

The quadratic equation is $$ (a^2+b^2)x^2 - 2b(a+c)x + (b^2+c^2) = 0 $$ $a$, $b$, and $c$ are non-zero, real and distinct numbers. This equation has non-zero real roots $(D \geq 0)$. One of the roots of the above equation is also the root of which equation from the following:

A) $$(b^2-c^2)x^2 + 2a(b-c)x - (a^2) = 0$$ B) $$(b^2+c^2)x^2 - 2a(b+c)x + (a^2) = 0$$ C) $$(a^2)x^2 + a(c-b)x - (bc)=0$$ D) $$(a^2)x^2 - a(b-c)x + (bc)=0$$

Using $D \geq 0$, I got $b^2=ac$, and substituting the value of $b^2$ in the original equation, I get : $$ax^2 - 2bx + c = 0$$

How should I proceed further?

Answer 1

As an alternative to the fine method given in the comments, as you noticed the determinant for the given equation is

$$\Delta =-4(ac-b^2)^2 \ge 0 \implies b^2=ac$$

which leads to

$$(a+c)(ax^2-2\sqrt{ac}x+c)=0 \implies ax^2-2\sqrt{ac}x+c=0 \implies (\sqrt a x-\sqrt c)^2=0$$

that is $$x=\frac{\sqrt c}{\sqrt a}=\frac b a=\frac c b$$

from which we see that only case C is compatible.

Answer 2

Using $\,D \geq 0\,$, I got $\,b^2=ac\,$

Given this is as a multiple choice question, it is enough to check particular cases for elimination, and it turns out that it's enough to check one case here. The smallest distinct, positive integers satisfying $b^2=ac$ are $a=1,b=2,c=4$. In that case the original equation has the double root $\,x=2\,$, and the only given choice that has $\,2\,$ as a root is $\,C\,$, so the answer must be $\,C\,$.

$$ \begin{matrix} & & a=1,b=2,c=4 \\ &(a^2+b^2)x^2 - 2b(a+c)x + (b^2+c^2) = 0 & 5 x^2 - 20 x + 20 = \color{red}{5 (x - 2)^2 = 0} \\ A) &(b^2-c^2)x^2 + 2a(b-c)x - a^2 = 0 & -12 x^2 - 4 x - 1 = 0 \\ B) &(b^2+c^2)x^2 - 2a(b+c)x + a^2 = 0 & 20 x^2 - 12 x + 1 = 0 \\ \color{red}{C)} &a^2x^2 + a(c-b)x - bc = 0 & x^2 + 2 x - 8 = \color{red}{(x-2)(x+4) = 0} \\ D) &a^2x^2 - a(b-c)x + bc = 0 & x^2 + 2 x + 8 = 0 \end{matrix} $$

Answer 3

You have some good answers already (one is just a comment but should probably have been an answer). Here's a slightly different path to the solution.

Once you have found that $b^2 = ac$ as observed in the question, you know that $a$, $b$, and $c$ are in geometric progression. So write $b = ar$ and $c = ar^2$. Making these substitutions in the original equation,

$$ (a^2+a^2r^2)x^2 - 2ar(a+ar^2)x + (a^2r^2+a^2r^4) = 0. $$

Since $a \neq 0$ we can divide all terms by $a^2$:

$$ (1+r^2)x^2 - 2r(1+r^2)x + (r^2+r^4) = 0. $$

Since $1 + r^2 \neq 0$ we can divide by $1 + r^2$:

$$ x^2 - 2rx + r^2 = 0. $$

This equation has a double root, $x = r$. That is, as in the other answers, $x = \dfrac ba = \dfrac cb$, and only C can be true.