Solve sqrt{5^2-x^2}=sqrt{6^2-(4-x)^2}

$$\sqrt { 5 ^ { 2 } - x ^ { 2 } } = \sqrt { 6 ^ { 2 } - ( 4 - x ) ^ { 2 } }$$

Solve for x

$$x=\frac{5}{8}=0.625$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{5^{2}-x^{2}}\right)^{2}=\left(\sqrt{6^{2}-\left(4-x\right)^{2}}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$\left(\sqrt{25-x^{2}}\right)^{2}=\left(\sqrt{6^{2}-\left(4-x\right)^{2}}\right)^{2}$$

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

$$25-x^{2}=\left(\sqrt{6^{2}-\left(4-x\right)^{2}}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$25-x^{2}=\left(\sqrt{36-\left(4-x\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-x\right)^{2}.

$$25-x^{2}=\left(\sqrt{36-\left(16-8x+x^{2}\right)}\right)^{2}$$

To find the opposite of 16-8x+x^{2}, find the opposite of each term.

$$25-x^{2}=\left(\sqrt{36-16+8x-x^{2}}\right)^{2}$$

Subtract 16 from 36 to get 20.

$$25-x^{2}=\left(\sqrt{20+8x-x^{2}}\right)^{2}$$

Calculate \sqrt{20+8x-x^{2}} to the power of 2 and get 20+8x-x^{2}.

$$25-x^{2}=20+8x-x^{2}$$

Subtract 8x from both sides.

$$25-x^{2}-8x=20-x^{2}$$

Add x^{2} to both sides.

$$25-x^{2}-8x+x^{2}=20$$

Combine -x^{2} and x^{2} to get 0.

$$25-8x=20$$

Subtract 25 from both sides.

$$-8x=20-25$$

Subtract 25 from 20 to get -5.

$$-8x=-5$$

Divide both sides by -8.

$$x=\frac{-5}{-8}$$

Fraction \frac{-5}{-8} can be simplified to \frac{5}{8} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{5}{8}$$

Substitute \frac{5}{8} for x in the equation \sqrt{5^{2}-x^{2}}=\sqrt{6^{2}-\left(4-x\right)^{2}}.

$$\sqrt{5^{2}-\left(\frac{5}{8}\right)^{2}}=\sqrt{6^{2}-\left(4-\frac{5}{8}\right)^{2}}$$

Simplify. The value x=\frac{5}{8} satisfies the equation.

$$\frac{15}{8}\times 7^{\frac{1}{2}}=\frac{15}{8}\times 7^{\frac{1}{2}}$$

Equation \sqrt{25-x^{2}}=\sqrt{36-\left(4-x\right)^{2}} has a unique solution.

$$x=\frac{5}{8}$$