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x\left(40x-8\right)=0
Factor out x.
x=0 x=\frac{1}{5}
To find equation solutions, solve x=0 and 40x-8=0.
40x^{2}-8x=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{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 40}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 40 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-8\right)±8}{2\times 40}
Take the square root of \left(-8\right)^{2}.
x=\frac{8±8}{2\times 40}
The opposite of -8 is 8.
x=\frac{8±8}{80}
Multiply 2 times 40.
x=\frac{16}{80}
Now solve the equation x=\frac{8±8}{80} when ± is plus. Add 8 to 8.
x=\frac{1}{5}
Reduce the fraction \frac{16}{80} to lowest terms by extracting and canceling out 16.
x=\frac{0}{80}
Now solve the equation x=\frac{8±8}{80} when ± is minus. Subtract 8 from 8.
x=0
Divide 0 by 80.
x=\frac{1}{5} x=0
The equation is now solved.
40x^{2}-8x=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{40x^{2}-8x}{40}=\frac{0}{40}
Divide both sides by 40.
x^{2}+\left(-\frac{8}{40}\right)x=\frac{0}{40}
Dividing by 40 undoes the multiplication by 40.
x^{2}-\frac{1}{5}x=\frac{0}{40}
Reduce the fraction \frac{-8}{40} to lowest terms by extracting and canceling out 8.
x^{2}-\frac{1}{5}x=0
Divide 0 by 40.
x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\left(-\frac{1}{10}\right)^{2}
Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{100}
Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.
\left(x-\frac{1}{10}\right)^{2}=\frac{1}{100}
Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. 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-\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}
Take the square root of both sides of the equation.
x-\frac{1}{10}=\frac{1}{10} x-\frac{1}{10}=-\frac{1}{10}
Simplify.
x=\frac{1}{5} x=0
Add \frac{1}{10} to both sides of the equation.