Solve (x)=sqrt{2+x-x^2}

$$( x ) = \sqrt { 2 + x - x ^ { 2 } }$$

Solve for x

$$x = \frac{\sqrt{17} + 1}{4} \approx 1.280776406$$

Solution Steps

Square both sides of the equation.

$$x^{2}=\left(\sqrt{2+x-x^{2}}\right)^{2}$$

Calculate \sqrt{2+x-x^{2}} to the power of 2 and get 2+x-x^{2}.

$$x^{2}=2+x-x^{2}$$

Subtract 2 from both sides.

$$x^{2}-2=x-x^{2}$$

Subtract x from both sides.

$$x^{2}-2-x=-x^{2}$$

Add x^{2} to both sides.

$$x^{2}-2-x+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-2-x=0$$

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.

$$2x^{2}-x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-2\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+16}}{2\times 2}$$

Add 1 to 16.

$$x=\frac{-\left(-1\right)±\sqrt{17}}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{17}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{17}}{4}$$

Now solve the equation x=\frac{1±\sqrt{17}}{4} when ± is plus. Add 1 to \sqrt{17}.

$$x=\frac{\sqrt{17}+1}{4}$$

Now solve the equation x=\frac{1±\sqrt{17}}{4} when ± is minus. Subtract \sqrt{17} from 1.

$$x=\frac{1-\sqrt{17}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+1}{4} x=\frac{1-\sqrt{17}}{4}$$

Substitute \frac{\sqrt{17}+1}{4} for x in the equation x=\sqrt{2+x-x^{2}}.

$$\frac{\sqrt{17}+1}{4}=\sqrt{2+\frac{\sqrt{17}+1}{4}-\left(\frac{\sqrt{17}+1}{4}\right)^{2}}$$

Simplify. The value x=\frac{\sqrt{17}+1}{4} satisfies the equation.

$$\frac{1}{4}\times 17^{\frac{1}{2}}+\frac{1}{4}=\frac{1}{4}\times 17^{\frac{1}{2}}+\frac{1}{4}$$

Substitute \frac{1-\sqrt{17}}{4} for x in the equation x=\sqrt{2+x-x^{2}}.

$$\frac{1-\sqrt{17}}{4}=\sqrt{2+\frac{1-\sqrt{17}}{4}-\left(\frac{1-\sqrt{17}}{4}\right)^{2}}$$

Simplify. The value x=\frac{1-\sqrt{17}}{4} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{4}-\frac{1}{4}\times 17^{\frac{1}{2}}=\frac{1}{4}\times 17^{\frac{1}{2}}-\frac{1}{4}$$

Equation x=\sqrt{2+x-x^{2}} has a unique solution.

$$x=\frac{\sqrt{17}+1}{4}$$