Solve sqrt{x^2+3^2}=sqrt{9+(2-x)^2}+frac{sqrt{10}}{2}

$$\sqrt { x ^ { 2 } + 3 ^ { 2 } } = \sqrt { 9 + ( 2 - x ) ^ { 2 } } + \frac { \sqrt { 10 } } { 2 }$$

Solve for x

$$x=\frac{5\sqrt{10}}{4}+1\approx 4.952847075$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}+3^{2}}\right)^{2}=\left(\sqrt{9+\left(2-x\right)^{2}}+\frac{\sqrt{10}}{2}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\left(\sqrt{x^{2}+9}\right)^{2}=\left(\sqrt{9+\left(2-x\right)^{2}}+\frac{\sqrt{10}}{2}\right)^{2}$$

Calculate \sqrt{x^{2}+9} to the power of 2 and get x^{2}+9.

$$x^{2}+9=\left(\sqrt{9+\left(2-x\right)^{2}}+\frac{\sqrt{10}}{2}\right)^{2}$$

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

$$x^{2}+9=\left(\sqrt{9+4-4x+x^{2}}+\frac{\sqrt{10}}{2}\right)^{2}$$

Add 9 and 4 to get 13.

$$x^{2}+9=\left(\sqrt{13-4x+x^{2}}+\frac{\sqrt{10}}{2}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{13-4x+x^{2}}+\frac{\sqrt{10}}{2}\right)^{2}.

$$x^{2}+9=\left(\sqrt{13-4x+x^{2}}\right)^{2}+2\sqrt{13-4x+x^{2}}\times \frac{\sqrt{10}}{2}+\left(\frac{\sqrt{10}}{2}\right)^{2}$$

Calculate \sqrt{13-4x+x^{2}} to the power of 2 and get 13-4x+x^{2}.

$$x^{2}+9=13-4x+x^{2}+2\sqrt{13-4x+x^{2}}\times \frac{\sqrt{10}}{2}+\left(\frac{\sqrt{10}}{2}\right)^{2}$$

Express 2\times \frac{\sqrt{10}}{2} as a single fraction.

$$x^{2}+9=13-4x+x^{2}+\frac{2\sqrt{10}}{2}\sqrt{13-4x+x^{2}}+\left(\frac{\sqrt{10}}{2}\right)^{2}$$

Cancel out 2 and 2.

$$x^{2}+9=13-4x+x^{2}+\sqrt{10}\sqrt{13-4x+x^{2}}+\left(\frac{\sqrt{10}}{2}\right)^{2}$$

To raise \frac{\sqrt{10}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$x^{2}+9=13-4x+x^{2}+\sqrt{10}\sqrt{13-4x+x^{2}}+\frac{\left(\sqrt{10}\right)^{2}}{2^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 13-4x+x^{2} times \frac{2^{2}}{2^{2}}.

$$x^{2}+9=\frac{\left(13-4x+x^{2}\right)\times 2^{2}}{2^{2}}+\sqrt{10}\sqrt{13-4x+x^{2}}+\frac{\left(\sqrt{10}\right)^{2}}{2^{2}}$$

Since \frac{\left(13-4x+x^{2}\right)\times 2^{2}}{2^{2}} and \frac{\left(\sqrt{10}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}+9=\frac{\left(13-4x+x^{2}\right)\times 2^{2}+\left(\sqrt{10}\right)^{2}}{2^{2}}+\sqrt{10}\sqrt{13-4x+x^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -4x+x^{2} times \frac{2^{2}}{2^{2}}.

$$x^{2}+9=13+\frac{\left(-4x+x^{2}\right)\times 2^{2}}{2^{2}}+\sqrt{10}\sqrt{13-4x+x^{2}}+\frac{\left(\sqrt{10}\right)^{2}}{2^{2}}$$

Since \frac{\left(-4x+x^{2}\right)\times 2^{2}}{2^{2}} and \frac{\left(\sqrt{10}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}+9=13+\frac{\left(-4x+x^{2}\right)\times 2^{2}+\left(\sqrt{10}\right)^{2}}{2^{2}}+\sqrt{10}\sqrt{13-4x+x^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2} times \frac{2^{2}}{2^{2}}.

$$x^{2}+9=13-4x+\frac{x^{2}\times 2^{2}}{2^{2}}+\sqrt{10}\sqrt{13-4x+x^{2}}+\frac{\left(\sqrt{10}\right)^{2}}{2^{2}}$$

Since \frac{x^{2}\times 2^{2}}{2^{2}} and \frac{\left(\sqrt{10}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$x^{2}+9=13-4x+\frac{x^{2}\times 2^{2}+\left(\sqrt{10}\right)^{2}}{2^{2}}+\sqrt{10}\sqrt{13-4x+x^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}+9=13-4x+\frac{x^{2}\times 4+\left(\sqrt{10}\right)^{2}}{2^{2}}+\sqrt{10}\sqrt{13-4x+x^{2}}$$

The square of \sqrt{10} is 10.

$$x^{2}+9=13-4x+\frac{x^{2}\times 4+10}{2^{2}}+\sqrt{10}\sqrt{13-4x+x^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}+9=13-4x+\frac{x^{2}\times 4+10}{4}+\sqrt{10}\sqrt{13-4x+x^{2}}$$

Divide each term of x^{2}\times 4+10 by 4 to get x^{2}+\frac{5}{2}.

$$x^{2}+9=13-4x+x^{2}+\frac{5}{2}+\sqrt{10}\sqrt{13-4x+x^{2}}$$

Add 13 and \frac{5}{2} to get \frac{31}{2}.

$$x^{2}+9=\frac{31}{2}-4x+x^{2}+\sqrt{10}\sqrt{13-4x+x^{2}}$$

Subtract \frac{31}{2}-4x+x^{2} from both sides of the equation.

$$x^{2}+9-\left(\frac{31}{2}-4x+x^{2}\right)=\sqrt{10}\sqrt{13-4x+x^{2}}$$

To find the opposite of \frac{31}{2}-4x+x^{2}, find the opposite of each term.

$$x^{2}+9-\frac{31}{2}+4x-x^{2}=\sqrt{10}\sqrt{13-4x+x^{2}}$$

Subtract \frac{31}{2} from 9 to get -\frac{13}{2}.

$$x^{2}-\frac{13}{2}+4x-x^{2}=\sqrt{10}\sqrt{13-4x+x^{2}}$$

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

$$-\frac{13}{2}+4x=\sqrt{10}\sqrt{13-4x+x^{2}}$$

Square both sides of the equation.

$$\left(-\frac{13}{2}+4x\right)^{2}=\left(\sqrt{10}\sqrt{13-4x+x^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\frac{13}{2}+4x\right)^{2}.

$$\frac{169}{4}-52x+16x^{2}=\left(\sqrt{10}\sqrt{13-4x+x^{2}}\right)^{2}$$

Expand \left(\sqrt{10}\sqrt{13-4x+x^{2}}\right)^{2}.

$$\frac{169}{4}-52x+16x^{2}=\left(\sqrt{10}\right)^{2}\left(\sqrt{13-4x+x^{2}}\right)^{2}$$

The square of \sqrt{10} is 10.

$$\frac{169}{4}-52x+16x^{2}=10\left(\sqrt{13-4x+x^{2}}\right)^{2}$$

Calculate \sqrt{13-4x+x^{2}} to the power of 2 and get 13-4x+x^{2}.

$$\frac{169}{4}-52x+16x^{2}=10\left(13-4x+x^{2}\right)$$

Use the distributive property to multiply 10 by 13-4x+x^{2}.

$$\frac{169}{4}-52x+16x^{2}=130-40x+10x^{2}$$

Subtract 130 from both sides.

$$\frac{169}{4}-52x+16x^{2}-130=-40x+10x^{2}$$

Subtract 130 from \frac{169}{4} to get -\frac{351}{4}.

$$-\frac{351}{4}-52x+16x^{2}=-40x+10x^{2}$$

Add 40x to both sides.

$$-\frac{351}{4}-52x+16x^{2}+40x=10x^{2}$$

Combine -52x and 40x to get -12x.

$$-\frac{351}{4}-12x+16x^{2}=10x^{2}$$

Subtract 10x^{2} from both sides.

$$-\frac{351}{4}-12x+16x^{2}-10x^{2}=0$$

Combine 16x^{2} and -10x^{2} to get 6x^{2}.

$$-\frac{351}{4}-12x+6x^{2}=0$$

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.

$$6x^{2}-12x-\frac{351}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -12 for b, and -\frac{351}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 6\left(-\frac{351}{4}\right)}}{2\times 6}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 6\left(-\frac{351}{4}\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-12\right)±\sqrt{144-24\left(-\frac{351}{4}\right)}}{2\times 6}$$

Multiply -24 times -\frac{351}{4}.

$$x=\frac{-\left(-12\right)±\sqrt{144+2106}}{2\times 6}$$

Add 144 to 2106.

$$x=\frac{-\left(-12\right)±\sqrt{2250}}{2\times 6}$$

Take the square root of 2250.

$$x=\frac{-\left(-12\right)±15\sqrt{10}}{2\times 6}$$

The opposite of -12 is 12.

$$x=\frac{12±15\sqrt{10}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{12±15\sqrt{10}}{12}$$

Now solve the equation x=\frac{12±15\sqrt{10}}{12} when ± is plus. Add 12 to 15\sqrt{10}.

$$x=\frac{15\sqrt{10}+12}{12}$$

Divide 12+15\sqrt{10} by 12.

$$x=\frac{5\sqrt{10}}{4}+1$$

Now solve the equation x=\frac{12±15\sqrt{10}}{12} when ± is minus. Subtract 15\sqrt{10} from 12.

$$x=\frac{12-15\sqrt{10}}{12}$$

Divide 12-15\sqrt{10} by 12.

$$x=-\frac{5\sqrt{10}}{4}+1$$

The equation is now solved.

$$x=\frac{5\sqrt{10}}{4}+1 x=-\frac{5\sqrt{10}}{4}+1$$

Substitute \frac{5\sqrt{10}}{4}+1 for x in the equation \sqrt{x^{2}+3^{2}}=\sqrt{9+\left(2-x\right)^{2}}+\frac{\sqrt{10}}{2}.

$$\sqrt{\left(\frac{5\sqrt{10}}{4}+1\right)^{2}+3^{2}}=\sqrt{9+\left(2-\left(\frac{5\sqrt{10}}{4}+1\right)\right)^{2}}+\frac{\sqrt{10}}{2}$$

Simplify. The value x=\frac{5\sqrt{10}}{4}+1 satisfies the equation.

$$5+\frac{1}{4}\times 10^{\frac{1}{2}}=5+\frac{1}{4}\times 10^{\frac{1}{2}}$$

Substitute -\frac{5\sqrt{10}}{4}+1 for x in the equation \sqrt{x^{2}+3^{2}}=\sqrt{9+\left(2-x\right)^{2}}+\frac{\sqrt{10}}{2}.

$$\sqrt{\left(-\frac{5\sqrt{10}}{4}+1\right)^{2}+3^{2}}=\sqrt{9+\left(2-\left(-\frac{5\sqrt{10}}{4}+1\right)\right)^{2}}+\frac{\sqrt{10}}{2}$$

Simplify. The value x=-\frac{5\sqrt{10}}{4}+1 does not satisfy the equation.

$$5-\frac{1}{4}\times 10^{\frac{1}{2}}=5+\frac{3}{4}\times 10^{\frac{1}{2}}$$

Substitute \frac{5\sqrt{10}}{4}+1 for x in the equation \sqrt{x^{2}+3^{2}}=\sqrt{9+\left(2-x\right)^{2}}+\frac{\sqrt{10}}{2}.

$$\sqrt{\left(\frac{5\sqrt{10}}{4}+1\right)^{2}+3^{2}}=\sqrt{9+\left(2-\left(\frac{5\sqrt{10}}{4}+1\right)\right)^{2}}+\frac{\sqrt{10}}{2}$$

Simplify. The value x=\frac{5\sqrt{10}}{4}+1 satisfies the equation.

$$5+\frac{1}{4}\times 10^{\frac{1}{2}}=5+\frac{1}{4}\times 10^{\frac{1}{2}}$$

Equation \sqrt{x^{2}+9}=\sqrt{\left(2-x\right)^{2}+9}+\frac{\sqrt{10}}{2} has a unique solution.

$$x=\frac{5\sqrt{10}}{4}+1$$