Question 595421: Factoring Quadratic formulas- Is this correct thus far?
5+2x^2=8x
2x^2-8x+5=0
a=2 b=-8 c=5
ac=2(5)=10 b=-8
Then i should find factors that multiply to give 10 & add to give -8 right ?
i only came up with the factors -5x(times)-2
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Factoring Quadratic formulas- Is this correct thus far?
5+2x^2=8x
2x^2-8x+5=0
a=2 b=-8 c=5
ac=2(5)=10 b=-8
Then i should find factors that multiply to give 10 & add to give -8 right ?
i only came up with the factors -5x(times)-2
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None of your operations are incorrect.
But, we don't factor quadratic formulas.
And, there is no quadratic formula.
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We factor the polynomials in order to solve the quadratic equation, if possible.
2x^2-8x+5=0
In this case, it can't be factored with integers, so we use the quadratic equation to solve it.
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| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=24 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 3.22474487139159, 0.775255128608411.
Here's your graph:
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Tho the solver says it can be factored, it's not using integers.
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