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