Solve for x
x = \frac{\sqrt{5} + 3}{2} \approx 2.618033989
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-\sqrt{x}=1-x
Subtract x from both sides of the equation.
\left(-\sqrt{x}\right)^{2}=\left(1-x\right)^{2}
Square both sides of the equation.
\left(-1\right)^{2}\left(\sqrt{x}\right)^{2}=\left(1-x\right)^{2}
Expand \left(-\sqrt{x}\right)^{2}.
1\left(\sqrt{x}\right)^{2}=\left(1-x\right)^{2}
Calculate -1 to the power of 2 and get 1.
1x=\left(1-x\right)^{2}
Calculate \sqrt{x} to the power of 2 and get x.
1x=1-2x+x^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.
x=x^{2}-2x+1
Reorder the terms.
x-x^{2}=-2x+1
Subtract x^{2} from both sides.
x-x^{2}+2x=1
Add 2x to both sides.
3x-x^{2}=1
Combine x and 2x to get 3x.
3x-x^{2}-1=0
Subtract 1 from both sides.
-x^{2}+3x-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{-3±\sqrt{3^{2}-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-1\right)}}{2\left(-1\right)}
Square 3.
x=\frac{-3±\sqrt{9+4\left(-1\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-3±\sqrt{9-4}}{2\left(-1\right)}
Multiply 4 times -1.
x=\frac{-3±\sqrt{5}}{2\left(-1\right)}
Add 9 to -4.
x=\frac{-3±\sqrt{5}}{-2}
Multiply 2 times -1.
x=\frac{\sqrt{5}-3}{-2}
Now solve the equation x=\frac{-3±\sqrt{5}}{-2} when ± is plus. Add -3 to \sqrt{5}.
x=\frac{3-\sqrt{5}}{2}
Divide -3+\sqrt{5} by -2.
x=\frac{-\sqrt{5}-3}{-2}
Now solve the equation x=\frac{-3±\sqrt{5}}{-2} when ± is minus. Subtract \sqrt{5} from -3.
x=\frac{\sqrt{5}+3}{2}
Divide -3-\sqrt{5} by -2.
x=\frac{3-\sqrt{5}}{2} x=\frac{\sqrt{5}+3}{2}
The equation is now solved.
\frac{3-\sqrt{5}}{2}-\sqrt{\frac{3-\sqrt{5}}{2}}=1
Substitute \frac{3-\sqrt{5}}{2} for x in the equation x-\sqrt{x}=1.
2-5^{\frac{1}{2}}=1
Simplify. The value x=\frac{3-\sqrt{5}}{2} does not satisfy the equation because the left and the right hand side have opposite signs.
\frac{\sqrt{5}+3}{2}-\sqrt{\frac{\sqrt{5}+3}{2}}=1
Substitute \frac{\sqrt{5}+3}{2} for x in the equation x-\sqrt{x}=1.
1=1
Simplify. The value x=\frac{\sqrt{5}+3}{2} satisfies the equation.
x=\frac{\sqrt{5}+3}{2}
Equation -\sqrt{x}=1-x has a unique solution.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}