SOLUTION: 2x^4-15x^3+18x^2=0 Solve for x

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Question 232291: 2x^4-15x^3+18x^2=0 Solve for x
Answer by eggsarecool(46) About Me  (Show Source):
You can put this solution on YOUR website!
Ok so first step is to factor out x%5E2
This would give you x%5E2%282x%5E2-15x%2B18%29=0
At this point you could factor the quadratic to become.
x%5E2%283-2x%29%286-x%29=0 At this point if one term becomes 0 the whole thing does since anything times 0 is 0.
Now solve +3-2x=0 add 2x to both sides.
+2x=3+ and now divide both sides by 2
x=3%2F2
For 6-x=0 add x to both sides
x=6
And for x%5E2=0 take the square root of both sides
+sqrt%28x%5E2%29=sqrt%280%29 since sqrt%28x%5E2%29=x
x=0
Your final answer for the problem would be x=+3%2F2%2C+6%2C+0
Now if you are not comfortable factoring the quadratic %282x%5E2-15x%2B18%29
You could use the quadratic formula instead x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
For the quadratic formula you take the numbers from %282x%5E2-15x%2B18%29
To give you +a=2 b=-15 c=18
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-15x%2B18+=+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=%28-15%29%5E2-4%2A2%2A18=81.

Discriminant d=81 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--15%2B-sqrt%28+81+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-15%29%2Bsqrt%28+81+%29%29%2F2%5C2+=+6
x%5B2%5D+=+%28-%28-15%29-sqrt%28+81+%29%29%2F2%5C2+=+1.5

Quadratic expression 2x%5E2%2B-15x%2B18 can be factored:
2x%5E2%2B-15x%2B18+=+2%28x-6%29%2A%28x-1.5%29
Again, the answer is: 6, 1.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-15%2Ax%2B18+%29


And 3%2F2=1.5 it is just has just been converted to a decimal by the computer.
So you would get x=3%2F2 x=1.5 from the quadratic formula, and then you would still get x=0 from x%5E2=0. Same final answer, just two different methods.