SOLUTION: Can someone explain how to find the zeroes of the following problem? f(x) = 2x^2-x+4

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Question 275259: Can someone explain how to find the zeroes of the following problem?
f(x) = 2x^2-x+4
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The zeros are the x values which make the equation 2x%5E2-x%2B4=0 true.




2x%5E2-x%2B4=0 Start with the given equation.


Notice that the quadratic 2x%5E2-x%2B4 is in the form of Ax%5E2%2BBx%2BC where A=2, B=-1, and C=4


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%28-1%29+%2B-+sqrt%28+%28-1%29%5E2-4%282%29%284%29+%29%29%2F%282%282%29%29 Plug in A=2, B=-1, and C=4


x+=+%281+%2B-+sqrt%28+%28-1%29%5E2-4%282%29%284%29+%29%29%2F%282%282%29%29 Negate -1 to get 1.


x+=+%281+%2B-+sqrt%28+1-4%282%29%284%29+%29%29%2F%282%282%29%29 Square -1 to get 1.


x+=+%281+%2B-+sqrt%28+1-32+%29%29%2F%282%282%29%29 Multiply 4%282%29%284%29 to get 32


x+=+%281+%2B-+sqrt%28+-31+%29%29%2F%282%282%29%29 Subtract 32 from 1 to get -31


x+=+%281+%2B-+sqrt%28+-31+%29%29%2F%284%29 Multiply 2 and 2 to get 4.


x+=+%281+%2B-+i%2Asqrt%2831%29%29%2F%284%29 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


Recall that 'i' is the imaginary number and i=sqrt%28-1%29


x+=+%281%2Bi%2Asqrt%2831%29%29%2F%284%29 or x+=+%281-i%2Asqrt%2831%29%29%2F%284%29 Break up the expression.


So the zeros are x+=+%281%2Bi%2Asqrt%2831%29%29%2F%284%29 or x+=+%281-i%2Asqrt%2831%29%29%2F%284%29


Note: if you've never seen imaginary numbers before, then the answer is simply "no solutions"