# Prove that $b^2 pr =q^2 ac$ using matrices

Let $i_1,i_2$ and $j_1,j_2$ be non-zero real roots of $ax^2+bx+c$ and $px^2+qx+r$ respectively, where a,p $\neq$0. If the system of equations $i_1y+i_2z=0$ and $j_1y+j_2z=0$ has a non-trivial solution, then prove that$b^2 pr= q^2 ac$.

I proved it as follows -

Let $i_1 = \frac{-b + \sqrt{b^2 -4ac}}{2a}$ and $i_2=\frac{-b - \sqrt{b^2 -4ac}}{2a}$

$j_1 = \frac{-q + \sqrt{q^2 -4pr}}{2p}$ and $j_2=\frac{-q - \sqrt{q^2 -4pr}}{2p}$

$i_1y+i_2z=0$ can be re-arranged to $$\frac{b}{2a} (y+z) = \frac{\sqrt{b^2 -4ac}}{2a}(y-z)$$ $$b(y+z)=\sqrt{b^2 -4ac}(y-z) -------------1$$ Similarly, $$q(y+z)=\sqrt{q^2 -4pr}(y-z) -------------2$$ can be obtained.

I can't think of a way to say that y$\neq$ z , but, $\frac{[1]}{[2]}$ will give the required relationship.

Is my method correct and is there a way to solve this using matrices?

I know the systems of equations can be written as -

$$\begin{pmatrix} a & b \\ p & q \end{pmatrix} \begin{pmatrix} x^2\\ x \end{pmatrix} = \begin{pmatrix} -c\\ -r \end{pmatrix}$$ and $$\begin{pmatrix} i_1 & i_2 \\ j_1 & j_2 \end{pmatrix} \begin{pmatrix} y\\ z \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}$$ but I do not know how to continue this method. Any help is appreciated.

• $y$ is allowed to be equal to $z$ but in that case $i_1 = -i_2$ and $j_1 = -j_2$ and so you must have that $b =0$ and $q=0$ so it is trivially true that $b^2pr = q^2ac$, since they're both equal to zero. – DavidButlerUofA Jul 9 '14 at 0:45

You can use a fact about matrices, namely that if $A$ is square and $A{\bf x}={\bf0}$ has a non-trivial solution then $\det A=0$. This gives $$i_1j_2=i_2j_1\ .$$ Now substitute in formulae for the roots and simplify: $$\bigl(-b+\sqrt{b^2-4ac}\bigr) \bigl(-q+\sqrt{\smash{q^2-4pr}\vphantom{b^2}}\bigr) =\bigl(-b-\sqrt{b^2-4ac}\bigr) \bigl(-q-\sqrt{\smash{q^2-4pr}\vphantom{b^2}}\bigr)$$ so $$-b\sqrt{\smash{q^2-4pr}\vphantom{b^2}}-q\sqrt{b^2-4ac} =b\sqrt{\smash{q^2-4pr}\vphantom{b^2}}+q\sqrt{b^2-4ac}\ .$$ Hence $$2b\sqrt{\smash{q^2-4pr}\vphantom{b^2}}=-2q\sqrt{b^2-4ac}\ ,\tag{*}$$ and squaring both sides completes the proof.

Note that if we arranged the roots differently in our first equation, then $(*)$ would become $$2b\sqrt{\smash{q^2-4pr}\vphantom{b^2}}=2q\sqrt{b^2-4ac}\ ,$$ but since we are going to square both sides, it doesn't change the result.

In your solution, you should consider two more cases. Note that $y,z$ form a non-trivial solution of the linear equations, so they are not both zero.

• If $y=z$ then $y\ne-z$, so $(1)$ and $(2)$ give $b=0$ and $q=0$.
• If $y=-z$ then $(1)$ and $(2)$ give $b^2=4ac$ and $q^2=4pr$.

In both cases the result follows easily.

I don't know about matrices, but here is an alternative proof:

Since we know the roots of the polynomials, we can factorise to get: $ax^2 + bx + c = a(x-i_1)(x-i_2)$ and $px^2 + qx + c = p(x-j_1)(x-j_2)$.

From this, we can find that $a(i_1 + i_2) = -b$, $ai_1 i_2 = c$, $p(j_1 + j_2) = -q$ and $pj_1 j_2 = r$.

Now consider the system $i_1 y + i_2 x = 0, j_1 y + j_2 z =0$, which has a nontrivial solution.

Suppose $i_1 = i_2 = 0$. Then the first equation is actually $ax^2$ so both $b$ and $c$ are zero. Hence $b^2pr = q^2ac$ since they are both zero.

Now suppose $i_1$ and $i_2$ are not both zero. In this case, a nontrivial solution only happens when $(j_1, j_2)$ is a multiple of $(i_1,i_2)$. So let $j_1 = k i_1$ and $j_2 = k i_2$ for some $k$.

Using the equations above: $$p(j_1 j_2) = r\\ p(k i_1 k i_2) = r\\ k^2 p i_1 i_2 = r\\ k^2 p a i_1 i_2 = ar\\ k^2 p c = ar$$ and also: $$p(j_1 + j_2 ) = -q\\ p(ki_1 + ki_2) = -q\\ kp(i_1 + i_2) = -q\\ kpa(i_1 +i_2) = -qa\\ kp(-b) = -qa\\ kpb = qa\\ k^2p^2b^2 = q^2 a^2\\ k^2p^2b^2c = q^2 a^2c\\ k^2 pc b^2p = q^2 a^2c\\ ar b^2p = q^2 a^2c\\ b^2pr = q^2 ac$$

(Note I have specifically avoided dividing by k, since I don't know if it is equal to zero.)

The relations between the roots and the coefficients are:

$b=-a(i_{1}+i_{2})$

$c=ai_{1}i_{2}$

$q=-p(j_{1}+j_{2})$

$r=pj_{1}j_{2}$

The condition for the system to have nontrivial solutions is that the corresponding matrix is noninvertible i.e. it's determinant is zero:

$i_{1}j_{2}-i_{2}j_{1}=0$

We have:

$b^{2}pr=a^{2}p^{2}j_{1}j_{2}(i_{1}+i_{2})^{2}$

and

$q^{2}ac=a^{2}p^{2}i_{1}i_{2}(j_{1}+j_{2})^{2}$

Hence, you have to show that:

$j_{1}j_{2}(i_{1}+i_{2})^{2}=i_{1}i_{2}(j_{1}+j_{2})^{2}$

under the above condition:

$i_{1}j_{2}=i_{2}j_{1}$