How do i approach ahead in this question

Question

Let $a,b \in\Bbb N$, $a$ is not equal to $b$ and the quadratic equations $(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0$ and $(b-1)x^2 -(b^2 + 2)x +b^2 +2b=0$ have a common root, then the value of $ab/5$ is

So what I did was,

I subtracted the two equations and got

$x=(a+b+2)/(a+b)$ I tried putting it in equation,it didn’t work,then i tried adding the equations and then put the value,still didn’t work.

I cant seem to figure out how to approach this problem.Can anybody help out?

Answer 1

Let $q_a(x)=(a-1)x^2-(a^2+2)x+a^2+2a$. Any root $r$ of $q_a(x)$ and of $q_b(x)$ is also a root of$$(b-1)q_a(x)-(a-1)q_b(x)=\left(-a^2b+a^2+a b^2+2 a-b^2-2b\right)(x-1).$$Therefore, $r=1$ or$$-a^2b+a^2+a b^2+2 a-b^2-2b=0.\tag1$$But you can't have $r=1$, because\begin{align}1\text{ is a root of }q_a(x)&\iff q_a(1)=0\\&\iff3a-3=0\\&\iff a=1.\end{align}So, if $1$ was a root of both $q_a(x)$ and $q_b(x)$, you would have $a=b=1$, but you are assuming that $a\ne b$.

If, on the other hand, you have $(1)$, then $a=b$ or $a=\frac{b+2}{b-1}=1+\frac3{b-1}$. Since $a,b\in\Bbb N$, this can only occur in two case: when $b=2$ (in which case $a=4$) and when $b=4$ (in which case $a=2$). In both cases, you have $\frac{ab}5=\frac85$.

Answer 2

Hint: $\;t=a,b$ are the two roots of the quadratic $\,(t-1)x^2-(t^2+2)x+t^2+2t=0\,$ for the value of $\,x\,$ which is the common root.

Answer 3

If you like using Vieta's formulas, here is the possible solution based on these formulas:

Assuming $x$ as a coefficient, you can observe that, $a$ and $b$ are root of the quadratic respect to the variable $u:$

$$u^2(1-x)+u(x^2+2)-(x^2+2x)=0, ~ x≠1$$

Then, using the Vieta's formulas, we have

$$\begin{align}&ab=\frac{x^2+2x}{x-1}\\ \implies &x^2+x(2-ab)+ab=0 \\ \implies &x_1x_2=ab\end{align}$$

Then, we also have a quadratic respect to the variable $x:$

$$(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0, ~ a≠1$$

Since $a≠0$, applying Vieta's formulas again, we get

$$\begin{align}&\frac{a^2+2a}{a-1}=ab\\ \implies &b=\frac{a+2}{a-1}\\ \implies &b=1+\frac{3}{a-1}\\ \implies &a\in\left\{2,4\right\}\\ \implies &ab=8 \\ \implies &\frac{ab}{5}=\frac 85. \end{align}$$