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x^{2}-8x=4
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^{2}-8x-4=4-4
Subtract 4 from both sides of the equation.
x^{2}-8x-4=0
Subtracting 4 from itself leaves 0.
x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-4\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-8\right)±\sqrt{64-4\left(-4\right)}}{2}
Square -8.
x=\frac{-\left(-8\right)±\sqrt{64+16}}{2}
Multiply -4 times -4.
x=\frac{-\left(-8\right)±\sqrt{80}}{2}
Add 64 to 16.
x=\frac{-\left(-8\right)±4\sqrt{5}}{2}
Take the square root of 80.
x=\frac{8±4\sqrt{5}}{2}
The opposite of -8 is 8.
x=\frac{4\sqrt{5}+8}{2}
Now solve the equation x=\frac{8±4\sqrt{5}}{2} when ± is plus. Add 8 to 4\sqrt{5}.
x=2\sqrt{5}+4
Divide 8+4\sqrt{5} by 2.
x=\frac{8-4\sqrt{5}}{2}
Now solve the equation x=\frac{8±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from 8.
x=4-2\sqrt{5}
Divide 8-4\sqrt{5} by 2.
x=2\sqrt{5}+4 x=4-2\sqrt{5}
The equation is now solved.
x^{2}-8x=4
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.
x^{2}-8x+\left(-4\right)^{2}=4+\left(-4\right)^{2}
Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-8x+16=4+16
Square -4.
x^{2}-8x+16=20
Add 4 to 16.
\left(x-4\right)^{2}=20
Factor x^{2}-8x+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-4\right)^{2}}=\sqrt{20}
Take the square root of both sides of the equation.
x-4=2\sqrt{5} x-4=-2\sqrt{5}
Simplify.
x=2\sqrt{5}+4 x=4-2\sqrt{5}
Add 4 to both sides of the equation.