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