1
$\begingroup$

Given $$(m-2)x^2-4x+3=3-x^2+2nx$$ Compute the value of $m$.

$$(m-2)x^2-4x+3=3-x^2+2nx \\mx^2+(-4m-4)x+7=-x^2+2nx+3 \\m=-1, n=0$$

The answer is $m=1,n=-2$, please tell me where did I go wrong

$\endgroup$
1
  • 3
    $\begingroup$ There is no linear term in $(m-2)x^2$, it's purely quadratic in $x$. $\endgroup$
    – lulu
    Commented Sep 23, 2018 at 13:38

1 Answer 1

Reset to default
2
$\begingroup$

Hint:

The coefficient of $x^2$ on the left of the equal sign is $(m-2)$ and on the right it is $-1$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .