Hi, I'm having a hard time dealing with this problem. It says "If (2m+1)(x^2+1)=(2m+5)x has equal roots, find the value of m"
How do I solve this problem?
Well to start I would distribute those and get:
$\displaystyle 2mx^2 + 2m + x^2 + 1 = 2mx + 5x$
$\displaystyle = (2m + 1)x^2 + (2m + 5)x + (2m + 1) = 0$ (Set it equal to 0 since we need to find the roots.
Apply the quadratic equation:
$\displaystyle \frac{-(2m + 5) \pm \sqrt{(2m + 5)^2 - 4(2m + 1)^2}}{2(2m + 1)}$
How ever, we need these roots to be equal to m, therefore we set this equation to m:
$\displaystyle \frac{-(2m + 5) \pm \sqrt{(2m + 5)^2 - 4(2m + 1)^2}}{2(2m + 1)} = m$
And from there, solve for m . Hope this helps, Chad.
hi , let's change it into the general form : ax^2 + bx + c = 0 first
$\displaystyle (2m+1)(x^2+1) = (2m+5)x $
$\displaystyle (2m+1)x^2 - (2m+5)x + (2m+1) =0$
Since the equation has two equal roots , the discriminant $\displaystyle b^2 - 4ac = 0$
$\displaystyle [-(2m+5)]^2 - 4(2m+1)(2m+1) = 0 $
$\displaystyle (2m+5)^2 - 2^2(2m+1)^2 = 0 $
$\displaystyle [(2m+5)-(2)(2m+1)] [(2m+5) + (2)(2m+1) ] =0 $
$\displaystyle ( -2m + 3)( 6m + 7) = 0 $
what's next ?