SOLUTION: how do you solve 3=2x^2=5x

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Question 558089: how do you solve 3=2x^2=5x
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
3=2x%5E2=5x is two equations, and there is no solution that will work for both.
You probably meant 3=2x%5E2-5x
From 3=2x%5E2-5x
subtracting 3 from both sides, we get
0=2x%5E2-5x-3 or 2x%5E2-5x-3=0
That is a quadratic equation.
ONE WAY TO SOLVE IT
One of the ways to solve this particular quadratic equation is by factoring. If you are skilled at factoring polynomials, you could figure out that
2x%5E2-5x-3=%282x%2B1%29%28x-3%29 and can write the equation as
%282x%2B1%29%28x-3%29=0
The solutions are values of x that make one of those factors zero.
From 2x%2B1=0 we get x=-1%2F2
and from x-3=0 we get x=3
ANOTHER WAY
A way that works for all quadratic equations is using the quadratic formula
For any equation of the form ax%5E2%2Bbx%2Bc=0 the solutions are given by
x = %28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
For 2x%5E2-5x-3=0 , a=2, b=-5 and c=-3
The quadratic formula gives you
x = %28-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2-4%2A2%2A%28-3%29+%29%29%2F%282%2A2%29+ = %285+%2B-+sqrt%28+25%2B24+%29%29%2F4+ = %285+%2B-+sqrt%2849%29%29%2F4+ = %285+%2B-+7%29%2F4+
That gives you x = %285%2B7%29%2F4 = 12%2F4 = 3%29%29%29+and%0D%0A%7B%7B%7Bx = %285-7%29%2F4 = -2%2F4 = -1%2F2
AND YET ANOTHER WAY
If you do not like using the quadratic formula, you can "complete the square." The idea behind the "complete the square" trick is changing your equation so that you have the square of a binomial on one side of the equal sign and a number on the other side.
From 2x%5E2-5x=3 , you look at 2x%5E2-5x
and figure out what to add to it to make the square of a binomial appear there.
It could go like this
2x%5E2-5x=3 --> x%5E2-%285%2F2%29x=3%2F2 (dividing both sides by 2)
x%5E2-%285%2F2%29x%2B25%2F16=3%2F2%2B25%2F16 adding 25/16 to both sides,
to make %28x-5%2F4%29%5E2 = x%5E2-2%285%2F4%29x%2B25%2F16 = x%5E2-%285%2F2%29x%2B25%2F16 appear on the left side of the equal sign.
Then, replacing %28x-5%2F4%29%5E2 for the equivalent expression x%5E2-%285%2F2%29x%2B25%2F16 , the equation transforms into
%28x-5%2F4%29%5E2 = 3%2F2%2B25%2F16 --> %28x-5%2F4%29%5E2 = 49%2F16 --> %28x-5%2F4%29%5E2 = %287%2F4%29%5E2
The solutions would be
x-5%2F4=7%2F4 --> x=5%2F4%2B7%2F4 --> x=12%2F4 --> x=3 and
x-5%2F4=-7%2F4 --> x=5%2F4-7%2F4 --> x=-2%2F4 --> x=-1%2F2