# Math Help - Solving a Quadratic Equation by Modifying the Coefficient

1. ## Solving a Quadratic Equation by Modifying the Coefficient

I'm working my way through Mathematics for the Nonmathematician and in the section on quadratic equations the author introduces a method of solving them that involves creating a new equation where the roots are the rots of the old one, each increased by one-half the coefficient of x. One of the exercise equations is:

$2x^2+8x+6=0$

but I don't know how to solve this due to the coefficient of $x^2$. Can anybody explain it to me?

EDIT: It's actually the pq formula that's used.

2. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

Hello!
do uk quadratic formula or pq formula? any of them?

3. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

Originally Posted by Petrus
Hello!
do uk quadratic formula or pq formula? any of them?
Do I know them? This book covers the pq formula (I think it's the same you're referring to). Actually I think I got my original post wrong and that's what is used to solve that quadratic. Either way I still can't figure out how to do it.

4. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

is factoring of the expression an accepted method? or do you refer to the factoring method as the pq method?

5. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

Originally Posted by MAX09
is factoring of the expression an accepted method? or do you refer to the factoring method as the pq method?
The book says to solve it using the formula:

$\sqrt{{p^2 \over 4} - q} - {p \over 2}$

6. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

i guess i am unfamiliar with this one... have no idea what p and q are.. good luck solving this one...

7. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

Hi!
we know the pq formula goosh idk how to copy this but i give u the link Resultat av Googles bildsökning efter http://img97.imageshack.us/img97/1020/namnls5zf.jpg

first we divide by 2 because x^2 need to be 1 in pq formula! so we got x^2+4x+3=0
now if we use pq formula
x1,2=-4/2+-sqrt((4/2)^2-3)
can u finish this im sorry i cant use latex code so its alot easy to see what i do :S

8. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

Thanks. I should have realised you can just divide the whole expression by 2. Oh well.

9. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

do u want me to solve this for u? or did u solve it by ur self?

10. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

$2x^2 + 8x + 6 = x^2 + 4x + 3 = 0$

$x = \sqrt{{4^2 \over 4} - 3} - {4 \over 2}$

$x = \sqrt{4 - 3} - 2$

$x = \sqrt{1} - 2$

$x = 1 - 2$

$x = -1$

and

$x = -\sqrt{{4^2 \over 4} - 3} - {4 \over 2}$

$x = -\sqrt{4 - 3} - 2$

$x = -\sqrt{1} - 2$

$x = -1 - 2$

$x = -3$

So $x = 1$ or $x = -3$

11. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

You are, I hope, aware that this is simply the quadratic formula?

-Dan

12. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

I'm aware that there is a quadratic formula but not that it's equivalent to this.

13. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

U got some wrong it should be (-4/2) (ik that it dont really mather but a strange teacher can give u less point for that misstake

14. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

You are given to solve:

$2x^2+8x+6=0$

I would divide through by 2:

$x^2+4x+3=0$

I think the author wants you to then write:

$(x-2)^2+4(x-2)+3=0$

This equation will have two roots which are both 2 more than the original equation.

$(x^2-4x+4)+4(x-2)+3=0$

$x^2-4x+4+4x-8+3=0$

$x^2=1$

$x=\pm1$

Thus, the roots of the original equation will be:

$x=-2\pm1$

$x=-3,\,-1$

15. ## Re: Solving a Quadratic Equation by Modifying the Coefficient

Tbh the pq formula is quadratic formula but more stricter because x^2 needs to be alone

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