Solving Quadratic Equation With Equal Roots

**saberteeth** · August 27th 2009, 12:07 AM

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?

**red_dog** · August 27th 2009, 12:14 AM

$(2m+1)x^2-(2m+5)x+2m+1=0$

If the roots are equal then the discriminant is 0.

Find the discriminant and solve the quadratic in respect to m.

**eXist** · August 27th 2009, 12:19 AM

Well to start I would distribute those and get:

$2mx^2 + 2m + x^2 + 1 = 2mx + 5x$

$= (2m + 1)x^2 + (2m + 5)x + (2m + 1) = 0$ (Set it equal to 0 since we need to find the roots.

Apply the quadratic equation:

$\frac{-(2m + 5) \pm \sqrt{(2m + 5)^2 - 4(2m + 1)^2}}{2(2m + 1)}$

How ever, we need these roots to be equal to m, therefore we set this equation to m:

$\frac{-(2m + 5) \pm \sqrt{(2m + 5)^2 - 4(2m + 1)^2}}{2(2m + 1)} = m$

And from there, solve for m

. Hope this helps, Chad.

**simplependulum** · August 27th 2009, 12:19 AM

Originally Posted by saberteeth

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?

hi , let's change it into the general form : ax^2 + bx + c = 0 first

$(2m+1)(x^2+1) = (2m+5)x$

$(2m+1)x^2 - (2m+5)x + (2m+1) =0$

Since the equation has two equal roots , the discriminant $b^2 - 4ac = 0$

$[-(2m+5)]^2 - 4(2m+1)(2m+1) = 0$

$(2m+5)^2 - 2^2(2m+1)^2 = 0$

$[(2m+5)-(2)(2m+1)] [(2m+5) + (2)(2m+1) ] =0$

$( -2m + 3)( 6m + 7) = 0$

what's next ?

**saberteeth** · August 27th 2009, 12:32 AM

thanx!

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