Question 642382
{{{(1/2)x^2+x=1}}}


{{{x^2+2x=2}}} Multiply EVERY term by the LCD 2 to clear out the fraction.


{{{x^2+2x-2=0}}}


Now use the quadratic formula to solve for x


Note: {{{x^2+2x-2}}} really looks like <img src="http://latex.codecogs.com/gif.latex?\large 1x^2 + 2x + -2" title="\large 1x^2 + 2x + -2" /> and it is in the form {{{ax^2+bx+c}}} where a = 1, b = 2 and c = -2



{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(2)+-sqrt((2)^2-4(1)(-2)))/(2(1))}}}


{{{x = (-2+-sqrt(4-(-8)))/(2)}}}


{{{x = (-2+-sqrt(4+8))/(2)}}}


{{{x = (-2+-sqrt(12))/2}}}


{{{x = (-2+sqrt(12))/2}}} or {{{x = (-2-sqrt(12))/2}}}


{{{x = (-2+2*sqrt(3))/2}}} or {{{x = (-2-2*sqrt(3))/2}}}


{{{x = -1+sqrt(3)}}} or {{{x = -1-sqrt(3)}}}


{{{x = 0.732050807568877}}} or {{{x = -2.73205080756888}}}


So the exact solutions are {{{x = -1+sqrt(3)}}} or {{{x = -1-sqrt(3)}}}


and the approximate solutions are {{{x = 0.732050807568877}}} or {{{x = -2.73205080756888}}}


These approximate solutions round to {{{x = 0.732}}} or {{{x = -2.732}}}


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Final Answer:


{{{x = 0.732}}} or {{{x = -2.732}}}

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