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Let $a,b \in N $ and $a \neq b$

If $(a-1)x^2-(a^2+2)x+a^2+2a=0$

and

$(b-1)x^2-(b^2+2)x+b^2+2b=0$

have a common root then the value of $ab$ is ?

I tried using Cramer's rule for common root but that did not simplify to anything and on subtracting the equations I again ended up with a quadratic, on solving the quadratic I got

$$x=\frac{a^2-b^2\pm\sqrt{(b^2-a^2)^2-4(a-b)(a^2-b^2+2a-2b)}}{2(a-b)}$$ which does not simplify either.

Any hints on how should I solve this.

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  • $\begingroup$ You are able to isolate $(a-b)^2$ inside the radical by grouping $\endgroup$
    – Blex
    Commented Sep 6, 2017 at 18:03
  • $\begingroup$ Please type out your expression rather than posting a picture. $\endgroup$
    – Théophile
    Commented Sep 6, 2017 at 18:06
  • $\begingroup$ Did the accepted post below allow you to reach an answer to the question in your post? $\endgroup$
    – Did
    Commented Sep 8, 2017 at 11:46
  • $\begingroup$ Yes, why do you ask ? $\endgroup$
    – Gem
    Commented Sep 8, 2017 at 11:47
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    $\begingroup$ Yeah. You do not say the context you were asked this in but if I were you, I would readily forget the "answer" you saw fit to accept and instead, give a look or two at the link in Robert's answer. Just my two cents. $\endgroup$
    – Did
    Commented Sep 8, 2017 at 14:20

2 Answers 2

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If $f(x)$ and $g(x)$ are the left sides of your two equations, the resultant of $f(x)$ and $g(x)$ is $3(ab-a-b-2)^2(a-b)^2$. This is $0$ if and only if the equations have a common root. Thus $$ab = a + b + 2$$
Write this as $$(a-1)(b-1) = 3$$ S Since $3$ is prime, if $a$ and $b$ are natural numbers we need $a-1$ and $b-1$ to be $1$ and $3$ (in either order). Thus one of $a$ and $b$ is $2$ and the other is $4$.

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  • $\begingroup$ What do you mean by resultant ? $\endgroup$
    – Gem
    Commented Sep 6, 2017 at 18:16
  • $\begingroup$ Link added..... $\endgroup$ Commented Sep 6, 2017 at 18:18
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solving the two quadratic equations we get $$x_1=\frac{a+2}{a-1},x_2=a$$ $$x_3=\frac{b+2}{b-1},x_4=b$$ if we have $$x_1=x_3$$ we have $$(a+2)(b-1)=(b+2)(a-1)$$ solving this we obtain $a=b$ can you solve the other cases?

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  • $\begingroup$ But how would I know which 2 are equal $\endgroup$
    – Gem
    Commented Sep 6, 2017 at 18:11
  • $\begingroup$ hm you must calculate all possibilities $\endgroup$ Commented Sep 6, 2017 at 18:15
  • $\begingroup$ $x_3=x_2$ and $x_1=x_4$ both give answers so which one should I accept $\endgroup$
    – Gem
    Commented Sep 6, 2017 at 18:26
  • $\begingroup$ which answers do you got? $\endgroup$ Commented Sep 6, 2017 at 18:31
  • $\begingroup$ a = 4 or 2 and b = 2 or 4 $\endgroup$
    – Gem
    Commented Sep 6, 2017 at 18:32

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