SOLUTION: Hello my name is Juan I'm in college, but I have a big problem I have 3 questions The question says Solve algebraically show all work put your answer on the line. For calculat

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Question 1023025: Hello my name is Juan I'm in college, but I have a big problem I have 3 questions
The question says Solve algebraically show all work put your answer on the line. For calculation and final answer, round to the nearest hundredths. Make sure your handwriting is legible to me. check your answers
solve write your solutions as ordered pairs
1) x^2+8x-12=0
I solved like this (x-6) (x-2) or (x+6)(x-2)
I have try to also solved like this

x^2+8x-12=0
x^2+8x=12
x(x+8)=12
x(x+8)-12=0/x
x+8-12=0
x-4=0
x=4 Now I don't know if this answer is correct please help me I have tried everything but I still cannot found the answer
the next question is
5x^2+19x=35
what I have tried is almost the same as the last one
5x^2+19x=35
5x(x+19)=35
x+19=35/5
x+19=7
x+19-7=0
x+12=0
x=-12
and the third question is
-6x^2-x+5=0
(-1)-6x^2-x+5=0
6x-x+5=0
(6x-1)(x+5) Please help me I have seriously tried everything


Found 2 solutions by Alan3354, addingup:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Hello my name is Juan I'm in college, but I have a big problem I have 3 questions
The question says Solve algebraically show all work put your answer on the line. For calculation and final answer, round to the nearest hundredths. Make sure your handwriting is legible to me. check your answers
solve write your solutions as ordered pairs
1) x^2+8x-12=0
I solved like this (x-6) (x-2) or (x+6)(x-2)
I have try to also solved like this
x^2+8x-12=0
x^2+8x=12
x(x+8)=12
x(x+8)-12=0/x
x+8-12=0
x-4=0
x=4 Now I don't know if this answer is correct please help me I have tried everything but I still cannot found the answer
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Your name is of no interest, it has no effect on the problem.
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1) x^2+8x-12=0
The quadratic cannot be factored --> use the quadratic formula:
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x^2+8x-12=0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B8x%2B-12+=+0) has the following solutons:

x%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=%288%29%5E2-4%2A1%2A-12=112.

Discriminant d=112 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-8%2B-sqrt%28+112+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%288%29%2Bsqrt%28+112+%29%29%2F2%5C1+=+1.29150262212918
x%5B2%5D+=+%28-%288%29-sqrt%28+112+%29%29%2F2%5C1+=+-9.29150262212918

Quadratic expression 1x%5E2%2B8x%2B-12 can be factored:
1x%5E2%2B8x%2B-12+=+%28x-1.29150262212918%29%2A%28x--9.29150262212918%29
Again, the answer is: 1.29150262212918, -9.29150262212918. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B8%2Ax%2B-12+%29

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Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
On 1), when you FOIL you have to end up with x^2+8x-12=0. But this number is irreducible. The most you can do is x(x+8) = 12 or x(x+8)-12 = 0 etc. which is how you have it on your second answer. However, since it's irreducible I don't see a way to write it as an ordered pair. Maybe I'm missing something.
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5x^2+19x-35 = 0
x^2+(19x)/5-7 = 0
x^2+(19 x)/5 = 7
x^2+(19 x)/5+361/100 = 1061/100
(x+19/10)^2 = 1061/100
x+19/10 = sqrt(1061)/10 or x+19/10 = -sqrt(1061)/10
Subtract 19/10 from both sides:
x = sqrt(1061)/10-19/10 or x+19/10 = -sqrt(1061)/10
x = sqrt(1061)/10-19/10 or x = -19/10-sqrt(1061)/10
Note: When you went from 5x(x+19)=35 to x+19=35/5 you ignored an x and the fact that 19x also has to be divided by 5, in other words, you would have:x(x+19)/5= 7 which is what I have (I have x^2(19x)/5 = 7).
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-6x^2-x+5= 0 Finally, something to work with!!
-(x+1) (6 x-5) = 0
Suerte en tus estudios.
J