Quadratic Identities - I Need Help Please.

**name101** · August 8th 2007, 04:36 AM

Find the Values of m, p and q for which $2x^2 - x -1$ is equivilent to $m(x+1)^2 + p(x+1) + q$ .

I don't know how to show this. I have Tried to expand the Question but i could not come to anything rational.

In this type of question do i use the quadratic formula?
$<br /> = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}<br />$

If so then.

Line 1:
$\frac{-(-1) \pm \sqrt{-1^2 - 4*2*-1}}{2*-1}$ is equivilent to $\frac{-(px+p) \pm \sqrt{(px-p)^2 - 4*m(x+1)*q}}{2*m(x+1)^2}$

Line 2:
$\frac{+1 \pm \sqrt{1 + 8}}{-2}$ is equivilent to $\frac{-px-p \pm \sqrt{px^2 - 2p^2 x - 4mqx - 4mq}}{2mx^2 + 4mx + 2m}$

Line 3:
$\frac{+1 \pm 3}{-2}$ is equivilent to $\frac{-px-p \pm \sqrt{px^2 - 2p^2 x - 4mqx - 4mq}}{2mx^2 + 4mx + 2m}$

This is where i am stuck.

I know to find the values for m, p and q. I need to do simutanious equations. But i do not know where to apply it. I also think that using the Quadratic formula is also incorrect.

Any help is appreciated. An explanation is even better.

Thank you for any help.

**topsquark** · August 8th 2007, 04:41 AM

Originally Posted by name101

Find the Values of m, p and q for which $2x^2 - x -1$ is equivilent to $m(x+1)^2 + p(x+1) + q$ .

$2x^2 - x - 1 = m(x + 1)^2 + p(x + 1) + q$

$2x^2 - x - 1 = mx^2 + 2mx + m + px + p + q$

$2x^2 - x - 1 = mx^2 + (2m + p)x + (m + p + q)$

Thus
$m = 2$
$2m + p = -1$
$m + p + q = -1$

You can solve for p and q.

-Dan

**name101** · August 8th 2007, 04:53 AM

$m = 2$
$2m + p = -1$
$m + p + q = -1$

then
$2 * 2 + p = -1$
$4 + p = -1$
$p = -5$

$m + p + q = -1$
$2 + (-5) + q = -1$
$-3 + q = -1$
$q = 2$

that part was easy.

for some reason that made a lot more sence.
thanks

so i take it the point of expanding out is to get
$mx^2 + b + c = 0$
even though it is expressed as
$mx^2 + (2m + p)x + (m + p + q) = 0$

**topsquark** · August 8th 2007, 05:00 AM

Originally Posted by name101

$m = 2$
$2m + p = -1$
$m + p + q = -1$

then
$2 * 2 + p = -1$
$4 + p = -1$
$p = -5$

$m + p + q = -1$
$2 + (-5) + q = -1$
$-3 + q = -1$
$q = 2$

that part was easy.

for some reason that made a lot more sence.
thanks

so i take it the point of expanding out is to get
$mx^2 + b + c = 0$
even though it is expressed as
$mx^2 + (2m + p)x + (m + p + q) = 0$

Yup. Whenever we have two polynomial expressions equal to each other the coefficients of like terms must be equal to each other.

-Dan

**name101** · August 9th 2007, 03:06 AM

Find the value $m$ in $x^2 + 2mx - 6 = 0$ if one of the roots is 2.

i know the answer is $m = 0.5$
But i dont understand why.

**name101** · August 9th 2007, 03:26 AM

Edit

**ticbol** · August 9th 2007, 03:36 AM

Originally Posted by name101

Find the value $m$ in $x^2 + 2mx - 6 = 0$ if one of the roots is 2.

i know the answer is $m = 0.5$
But i dont understand why.

Let a = the other root, so,
x^2 +2mx -6 = (x-2)(x-a)
x^2 +2mx -6 = x^2 -ax -2x +2a
x^2 +2mx -6 = x^2 -(a+2)x +2a

So, 2a = -6
a = -3

And, -(a+2) = 2m
So, -(-3 +2) = 2m
1 = 2m
m = 1/2 = 0.5 ----------------answer.

**topsquark** · August 9th 2007, 10:57 AM

Originally Posted by name101

Find the value $m$ in $x^2 + 2mx - 6 = 0$ if one of the roots is 2.

i know the answer is $m = 0.5$
But i dont understand why.

Please post new questions in a new thread.

-Dan

**name101** · August 10th 2007, 05:07 PM

sorry i thought since it is related. It can go in the same thread.

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