Solve sqrt{x+6}=4-{left(sqrt{2-x}right)}^2

$$\sqrt{ x+6 } = 4- { \left( \sqrt{ 2-x } \right) }^{ 2 }$$

Solve for x

$$x=\frac{\sqrt{17}-3}{2}\approx 0.561552813$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+6}\right)^{2}=\left(4-\left(\sqrt{2-x}\right)^{2}\right)^{2}$$

Calculate \sqrt{x+6} to the power of 2 and get x+6.

$$x+6=\left(4-\left(\sqrt{2-x}\right)^{2}\right)^{2}$$

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

$$x+6=\left(4-\left(2-x\right)\right)^{2}$$

To find the opposite of 2-x, find the opposite of each term.

$$x+6=\left(4-2+x\right)^{2}$$

Subtract 2 from 4 to get 2.

$$x+6=\left(2+x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+x\right)^{2}.

$$x+6=4+4x+x^{2}$$

Subtract 4 from both sides.

$$x+6-4=4x+x^{2}$$

Subtract 4 from 6 to get 2.

$$x+2=4x+x^{2}$$

Subtract 4x from both sides.

$$x+2-4x=x^{2}$$

Combine x and -4x to get -3x.

$$-3x+2=x^{2}$$

Subtract x^{2} from both sides.

$$-3x+2-x^{2}=0$$

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.

$$-x^{2}-3x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8}}{2\left(-1\right)}$$

Add 9 to 8.

$$x=\frac{-\left(-3\right)±\sqrt{17}}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{17}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±\sqrt{17}}{-2}$$

Now solve the equation x=\frac{3±\sqrt{17}}{-2} when ± is plus. Add 3 to \sqrt{17}.

$$x=\frac{\sqrt{17}+3}{-2}$$

Divide 3+\sqrt{17} by -2.

$$x=\frac{-\sqrt{17}-3}{2}$$

Now solve the equation x=\frac{3±\sqrt{17}}{-2} when ± is minus. Subtract \sqrt{17} from 3.

$$x=\frac{3-\sqrt{17}}{-2}$$

Divide 3-\sqrt{17} by -2.

$$x=\frac{\sqrt{17}-3}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{17}-3}{2} x=\frac{\sqrt{17}-3}{2}$$

Substitute \frac{-\sqrt{17}-3}{2} for x in the equation \sqrt{x+6}=4-\left(\sqrt{2-x}\right)^{2}.

$$\sqrt{\frac{-\sqrt{17}-3}{2}+6}=4-\left(\sqrt{2-\frac{-\sqrt{17}-3}{2}}\right)^{2}$$

Simplify. The value x=\frac{-\sqrt{17}-3}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\left(\frac{1}{2}-\frac{1}{2}\times 17^{\frac{1}{2}}\right)=\frac{1}{2}-\frac{1}{2}\times 17^{\frac{1}{2}}$$

Substitute \frac{\sqrt{17}-3}{2} for x in the equation \sqrt{x+6}=4-\left(\sqrt{2-x}\right)^{2}.

$$\sqrt{\frac{\sqrt{17}-3}{2}+6}=4-\left(\sqrt{2-\frac{\sqrt{17}-3}{2}}\right)^{2}$$

Simplify. The value x=\frac{\sqrt{17}-3}{2} satisfies the equation.

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

Equation \sqrt{x+6}=4-\left(\sqrt{2-x}\right)^{2} has a unique solution.

$$x=\frac{\sqrt{17}-3}{2}$$