1. ## Quadratic Identities - I Need Help Please.

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

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

If so then.

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

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

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

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 $\displaystyle 2x^2 - x -1$ is equivilent to $\displaystyle m(x+1)^2 + p(x+1) + q$.
$\displaystyle 2x^2 - x - 1 = m(x + 1)^2 + p(x + 1) + q$

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

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

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

You can solve for p and q.

-Dan

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

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

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

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
$\displaystyle mx^2 + b + c = 0$
even though it is expressed as
$\displaystyle mx^2 + (2m + p)x + (m + p + q) = 0$

4. Originally Posted by name101
$\displaystyle m = 2$
$\displaystyle 2m + p = -1$
$\displaystyle m + p + q = -1$

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

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

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
$\displaystyle mx^2 + b + c = 0$
even though it is expressed as
$\displaystyle mx^2 + (2m + p)x + (m + p + q) = 0$
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 $\displaystyle m$ in $\displaystyle x^2 + 2mx - 6 = 0$ if one of the roots is 2.

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

6. Edit

7. Originally Posted by name101
Find the value $\displaystyle m$ in $\displaystyle x^2 + 2mx - 6 = 0$ if one of the roots is 2.

i know the answer is $\displaystyle m = 0.5$
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 $\displaystyle m$ in $\displaystyle x^2 + 2mx - 6 = 0$ if one of the roots is 2.

i know the answer is $\displaystyle m = 0.5$
But i dont understand why.
Please post new questions in a new thread.

-Dan

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