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?
$
= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$

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.

2. 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

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

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
$mx^2 + (2m + p)x + (m + p + q) = 0$

4. 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

5. 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.

6. Edit

7. 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.

8. 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.