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Question Statement:-

If the equations $ax^2+2bx+c=0$ and $a_1x^2+2b_1x+c_1=0$ have one and only one root common, then prove that $b^2-ac$ and $b_1^2-a_1c_1$ are perfect squares.

Attempt at a solution:-

Let the common root be $\alpha$. Then $x=\alpha$ must satisfy both the quadratic equations. Hence, we have

$a\alpha^2+2b\alpha+c=0\tag{1}$ $a_1\alpha^2+2b_1\alpha+c_1=0\tag{2}$

On solving $(1)$ and $(2)$, we get

$$\dfrac{\alpha^2}{2bc_1-2b_1c}=\dfrac{-\alpha}{ac_1-a_1c}=\dfrac{1}{2ab_1-2ba_1}$$

From this, we get $$(a_1c-ac_1)^2=(2ab_1-2a_1b)(2bc_1-2b_1c)$$

On expanding and manipulating this I couldn't get anywhere close to what the question needs me to prove.

If you can think of an intuitive solution using graphs of the quadratic equations then it would be helpful too.

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user350331
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  • I suppose $a,b,c$ are integers? – Fimpellizzeri Nov 08 '16 at 14:50
  • @Fimpellizieri - Why is it needed to be supposed that $a,b,c$ are integers, for your reference the book does not mention so. – user350331 Nov 08 '16 at 16:23
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    @user350331 The equations $x^2-6x+5$ and $\sqrt\pi x^2 - 4\sqrt\pi x + 3\sqrt\pi$ both have the root $x=1$ and no other root in common, but $(2\sqrt\pi)^2 - \sqrt\pi \cdot 3\sqrt\pi = \pi$ is not a perfect square. Your book may mention several pages earlier that you will only be using integers in this chapter, or there may be some other indication of what is intended. – David K Nov 08 '16 at 16:41
  • @DavidK- Oh yeah it does tell "to consider the coefficients to be integers unless stated otherwise". Thanks for that I had been solving my questions without considering that and now only did I think about it. – user350331 Nov 08 '16 at 16:44

2 Answers2

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This answer assumes $a,b,c$ are integers (and of course, if they are rational, one can multiply both sides of the equation so that this holds). Write the equations as

$$a(x-r)(x-s)=ax^2-a(r+s)x+ars=0\\a_1(x-r)(x-t)=a_1x^2-a_1(r+t)x+a_1rt=0$$

Of course, $r \neq s,t$. These imply:

\begin{align} &-a(r+s)=2b&&ars=c\\ &-a_1(r+t)=2b_1&&a_1rt=c_1 \end{align}

Then:

\begin{align} b^2-ac&=\frac{a^2(r+s)^2}{4}-a^2rs\\ &=\frac{a^2}{4}(r^2+2rs+s^2-4rs)\\ &=\frac{a^2}{4}(r^2-2rs+s^2)\\ &={\left(\frac{a(r-s)}{2}\right)}^2 \end{align}

Notice that because $a(r+s)$ is even ($-2b)$, $a$ is even or $r,s$ are the same parity. Whatever the case, $a(r-s)$ will be even, so the fraction above is an integer and we are done.

The case ${b_1}^2-a_1c_1$ is treated similarly.

Fimpellizzeri
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$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b_1\pm\sqrt{b_1^2-4a_1c_1}}{2a_1}$$ $$\sqrt{b^2-4ac}+-\sqrt{b_1^2-4a_1c_1}\in Q$$ When you square it you will get $$\sqrt{(b^2-4ac)(b_1^2-4a_1c_1)}\in Q$$ So $$b^2-4ac=p^2m,\qquad b_1^2-4a_1c_1=q^2m$$ $$(p+-q)\sqrt m \in Q$$ $$\sqrt m \in Q$$

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Djura Marinkov
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