Solve sqrt{16-x^2+32+{left(9/5right)}^2}=x

$$\sqrt{ 16- { x }^{ 2 } +32+ { \left( \frac{ 9 }{ 5 } \right) }^{ 2 } } =x$$

Solve for x

$$x = \frac{\sqrt{2562}}{10} \approx 5.061620294$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{16-x^{2}+32+\left(\frac{9}{5}\right)^{2}}\right)^{2}=x^{2}$$

Add 16 and 32 to get 48.

$$\left(\sqrt{48-x^{2}+\left(\frac{9}{5}\right)^{2}}\right)^{2}=x^{2}$$

Calculate \frac{9}{5} to the power of 2 and get \frac{81}{25}.

$$\left(\sqrt{48-x^{2}+\frac{81}{25}}\right)^{2}=x^{2}$$

Add 48 and \frac{81}{25} to get \frac{1281}{25}.

$$\left(\sqrt{\frac{1281}{25}-x^{2}}\right)^{2}=x^{2}$$

Calculate \sqrt{\frac{1281}{25}-x^{2}} to the power of 2 and get \frac{1281}{25}-x^{2}.

$$\frac{1281}{25}-x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$\frac{1281}{25}-x^{2}-x^{2}=0$$

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

$$\frac{1281}{25}-2x^{2}=0$$

Subtract \frac{1281}{25} from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}=-\frac{1281}{25}$$

Divide both sides by -2.

$$x^{2}=\frac{-\frac{1281}{25}}{-2}$$

Express \frac{-\frac{1281}{25}}{-2} as a single fraction.

$$x^{2}=\frac{-1281}{25\left(-2\right)}$$

Multiply 25 and -2 to get -50.

$$x^{2}=\frac{-1281}{-50}$$

Fraction \frac{-1281}{-50} can be simplified to \frac{1281}{50} by removing the negative sign from both the numerator and the denominator.

$$x^{2}=\frac{1281}{50}$$

Take the square root of both sides of the equation.

$$x=\frac{\sqrt{2562}}{10} x=-\frac{\sqrt{2562}}{10}$$

Substitute \frac{\sqrt{2562}}{10} for x in the equation \sqrt{16-x^{2}+32+\left(\frac{9}{5}\right)^{2}}=x.

$$\sqrt{16-\left(\frac{\sqrt{2562}}{10}\right)^{2}+32+\left(\frac{9}{5}\right)^{2}}=\frac{\sqrt{2562}}{10}$$

Simplify. The value x=\frac{\sqrt{2562}}{10} satisfies the equation.

$$\frac{1}{10}\times 2562^{\frac{1}{2}}=\frac{1}{10}\times 2562^{\frac{1}{2}}$$

Substitute -\frac{\sqrt{2562}}{10} for x in the equation \sqrt{16-x^{2}+32+\left(\frac{9}{5}\right)^{2}}=x.

$$\sqrt{16-\left(-\frac{\sqrt{2562}}{10}\right)^{2}+32+\left(\frac{9}{5}\right)^{2}}=-\frac{\sqrt{2562}}{10}$$

Simplify. The value x=-\frac{\sqrt{2562}}{10} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{10}\times 2562^{\frac{1}{2}}=-\frac{1}{10}\times 2562^{\frac{1}{2}}$$

Equation \sqrt{\frac{1281}{25}-x^{2}}=x has a unique solution.

$$x=\frac{\sqrt{2562}}{10}$$