SOLUTION: solve as quaditic equations 4x^2=13x+12

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Question 87208: solve as quaditic equations
4x^2=13x+12
Found 2 solutions by Earlsdon, jim_thompson5910:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
First, write the equation in standard form:ax%5E2%2Bbx%2Bc+=+0
4x%5E2-13x-12+=+0 Solve by factoring.
%284x%2B3%29%28x-4%29+=+0 Apply the zero products principle.
4x%2B3+=+0 or x-4+=+0
Therefore:
4x+=+-3
x+=+-3%2F4
Or
x+=+4

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
4x%5E2=13x%2B12

4x%5E2-13x-12=0 Get everything to one side

Now use the quadratic formula to solve for x:

Starting with the general quadratic

ax%5E2%2Bbx%2Bc

the general form of the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve 4%2Ax%5E2-13%2Ax-12

x+=+%2813+%2B-+sqrt%28+%28-13%29%5E2-4%2A4%2A-12+%29%29%2F%282%2A4%29 Plug in a=4, b=-13, and c=-12



x+=+%2813+%2B-+sqrt%28+169-4%2A4%2A-12+%29%29%2F%282%2A4%29 Square -13 to get 169



x+=+%2813+%2B-+sqrt%28+169%2B192+%29%29%2F%282%2A4%29 Multiply -4%2A-12%2A4 to get 192



x+=+%2813+%2B-+sqrt%28+361+%29%29%2F%282%2A4%29 Combine like terms in the radicand (everything under the square root)



x+=+%2813+%2B-+19%29%2F%282%2A4%29 Simplify the square root



x+=+%2813+%2B-+19%29%2F8 Multiply 2 and 4 to get 8

So now the expression breaks down into two parts

x+=+%2813+%2B+19%29%2F8 or x+=+%2813+-+19%29%2F8

Lets look at the first part:

x=32%2F8 Add the terms in the numerator
x=4 Divide

So one answer is
x=4
Now lets look at the second part:

x=-6%2F8 Subtract the terms in the numerator
x=-3%2F4 Divide

So another answer is
x=-3%2F4

So our solutions are:
x=4 or x=-3%2F4

Notice when we graph 4%2Ax%5E2-13%2Ax-12 we get:

+graph%28+500%2C+500%2C+-13%2C+14%2C+-13%2C+14%2C4%2Ax%5E2%2B-13%2Ax%2B-12%29+

and we can see that the roots are x=4 and x=-3%2F4. This verifies our answer