1. ## solving quadratic relation by factoring

solve 2xsquared-7x-4=0 by factoring.
my first instinct is putting two brackets
(2x- )(x- )=0
unable to continue

2. Originally Posted by euclid2
solve 2xsquared-7x-4=0 by factoring.
my first instinct is putting two brackets
(2x- )(x- )=0
unable to continue
your first instinct is right, except for the signs. if we have two negatives, then the constant term would be positive, since a negative times a negative gives a positive, though two wrongs don't make a right.

so one should be plus and one minus to get the -4. now, you want to come up with two numbers that when multiplied gives you -4.

what are our options, we can have 1 times 4 or 2 times 2, with the proper signs attached. if you can't see which to use, you can go by trial and error.

lets do 1 and 4

(2x 1)(x 4)

can we get what we want from this?

well, the middle term is given by the product of the inner terms plus the product of the outer terms (remember FOIL)

multiply the outer terms, we get 8x
multiply the inner terms, we get 1x

can we make -7x with this? you bet we can, we got lucky. -8x + x = -7x. so this tells us how to put the signs in

(2x + 1)(x - 4) .......put the negative on the 4 so as to get -8x and the plus sign on the 1 to get the positive x.

so we have (2x + 1)(x - 4) = 0

can you continue? (remember, if we have two numbers being multiplied result in zero, then one number or the other must be zero)

3. ok, that makes much more sense
so;
(2x+1)(X-4)=0
2X=-1 X=4
X=-1/2
so x=-1/2 or 4
thanks alot, thats what i get using the quadratic formula as well

4. Originally Posted by euclid2
ok, that makes much more sense
so;
(2x+1)(X-4)=0
2X=-1 X=4
X=-1/2
so x=-1/2 or 4
thanks alot, thats what i get using the quadratic formula as well
yes, that's correct

5. As a side note, you can also use the quadratic formula to factor a quadratic equation instead of immediately putting two brackets and trying to figure out what numbers go where. I think this method is faster.

Factor $\displaystyle 2x^{2}-7x-4=0$

Using the quadratic formula: $\displaystyle x={\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}={\dfrac{-(-7)\pm\sqrt{(-7)^{2}-4(2)(-4)}}{2(2)}}$.

So $\displaystyle x={-\dfrac{1}{2}}$ or $\displaystyle x={4}$.

Take $\displaystyle x={-\dfrac{1}{2}}$ and multiply both sides by 2, then add 1 to both sides. This results in $\displaystyle (2x+1)=0$.

Now take $\displaystyle x={4}$ and subtract 4 from both sides to get $\displaystyle (x-4)=0$

So $\displaystyle (2x+1)(x-4)=0$

Hope this helps! =)