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\frac{1}{4}x^{2}+\frac{1}{2}x-2=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\times \frac{1}{4}\left(-2\right)}}{2\times \frac{1}{4}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, \frac{1}{2} for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\times \frac{1}{4}\left(-2\right)}}{2\times \frac{1}{4}}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-\left(-2\right)}}{2\times \frac{1}{4}}
Multiply -4 times \frac{1}{4}.
x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+2}}{2\times \frac{1}{4}}
Multiply -1 times -2.
x=\frac{-\frac{1}{2}±\sqrt{\frac{9}{4}}}{2\times \frac{1}{4}}
Add \frac{1}{4} to 2.
x=\frac{-\frac{1}{2}±\frac{3}{2}}{2\times \frac{1}{4}}
Take the square root of \frac{9}{4}.
x=\frac{-\frac{1}{2}±\frac{3}{2}}{\frac{1}{2}}
Multiply 2 times \frac{1}{4}.
x=\frac{1}{\frac{1}{2}}
Now solve the equation x=\frac{-\frac{1}{2}±\frac{3}{2}}{\frac{1}{2}} when ± is plus. Add -\frac{1}{2} to \frac{3}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=2
Divide 1 by \frac{1}{2} by multiplying 1 by the reciprocal of \frac{1}{2}.
x=-\frac{2}{\frac{1}{2}}
Now solve the equation x=\frac{-\frac{1}{2}±\frac{3}{2}}{\frac{1}{2}} when ± is minus. Subtract \frac{3}{2} from -\frac{1}{2} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
x=-4
Divide -2 by \frac{1}{2} by multiplying -2 by the reciprocal of \frac{1}{2}.
x=2 x=-4
The equation is now solved.
\frac{1}{4}x^{2}+\frac{1}{2}x-2=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
\frac{1}{4}x^{2}+\frac{1}{2}x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
\frac{1}{4}x^{2}+\frac{1}{2}x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
\frac{1}{4}x^{2}+\frac{1}{2}x=2
Subtract -2 from 0.
\frac{\frac{1}{4}x^{2}+\frac{1}{2}x}{\frac{1}{4}}=\frac{2}{\frac{1}{4}}
Multiply both sides by 4.
x^{2}+\frac{\frac{1}{2}}{\frac{1}{4}}x=\frac{2}{\frac{1}{4}}
Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.
x^{2}+2x=\frac{2}{\frac{1}{4}}
Divide \frac{1}{2} by \frac{1}{4} by multiplying \frac{1}{2} by the reciprocal of \frac{1}{4}.
x^{2}+2x=8
Divide 2 by \frac{1}{4} by multiplying 2 by the reciprocal of \frac{1}{4}.
x^{2}+2x+1^{2}=8+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=8+1
Square 1.
x^{2}+2x+1=9
Add 8 to 1.
\left(x+1\right)^{2}=9
Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+1\right)^{2}}=\sqrt{9}
Take the square root of both sides of the equation.
x+1=3 x+1=-3
Simplify.
x=2 x=-4
Subtract 1 from both sides of the equation.