SOLUTION: How do you solve the quadratic formula: all radicals should be simplified as far as possible, show work a^2-12a+35=0

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Question 297874: How do you solve the quadratic formula: all radicals should be simplified as far as possible, show work
a^2-12a+35=0
Found 2 solutions by checkley77, JBarnum:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
a^2-12a+35=0
(a-7)(a-5)=0
a-7=0
a=7 ans.
a-5=0
a=5 ans.

Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aa%5E2%2Bba%2Bc=0 (in our case 1a%5E2%2B-12a%2B35+=+0) has the following solutons:

a%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-12%29%5E2-4%2A1%2A35=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--12%2B-sqrt%28+4+%29%29%2F2%5Ca.

a%5B1%5D+=+%28-%28-12%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+7
a%5B2%5D+=+%28-%28-12%29-sqrt%28+4+%29%29%2F2%5C1+=+5

Quadratic expression 1a%5E2%2B-12a%2B35 can be factored:
1a%5E2%2B-12a%2B35+=+1%28a-7%29%2A%28a-5%29
Again, the answer is: 7, 5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-12%2Ax%2B35+%29