Solve 5^2+(frac{25sqrt{3}}{4}-2sqrt{3})^2=4^2+(frac{sqrt{3}sqrt{3}}{4}-b)^2

$$5 ^ { 2 } + ( \frac { 25 \sqrt { 3 } } { 4 } - 2 \sqrt { 3 } ) ^ { 2 } = 4 ^ { 2 } + ( \frac { \sqrt { 3 } \sqrt { 3 } } { 4 } - b ) ^ { 2 }$$

Solve for b

$$b=\frac{3-\sqrt{1011}}{4}\approx -7.199056548$$

$$b = \frac{\sqrt{1011} + 3}{4} \approx 8.699056548$$

Steps Using the Quadratic Formula

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$5^{2}+\left(\frac{25\sqrt{3}}{4}-2\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$25+\left(\frac{25\sqrt{3}}{4}-2\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Combine \frac{25\sqrt{3}}{4} and -2\sqrt{3} to get \frac{17}{4}\sqrt{3}.

$$25+\left(\frac{17}{4}\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Expand \left(\frac{17}{4}\sqrt{3}\right)^{2}.

$$25+\left(\frac{17}{4}\right)^{2}\left(\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Calculate \frac{17}{4} to the power of 2 and get \frac{289}{16}.

$$25+\frac{289}{16}\left(\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

The square of \sqrt{3} is 3.

$$25+\frac{289}{16}\times 3=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Multiply \frac{289}{16} and 3 to get \frac{867}{16}.

$$25+\frac{867}{16}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Add 25 and \frac{867}{16} to get \frac{1267}{16}.

$$\frac{1267}{16}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{1267}{16}=16+\left(\frac{3}{4}-b\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{4}-b\right)^{2}.

$$\frac{1267}{16}=16+\frac{9}{16}-\frac{3}{2}b+b^{2}$$

Add 16 and \frac{9}{16} to get \frac{265}{16}.

$$\frac{1267}{16}=\frac{265}{16}-\frac{3}{2}b+b^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{265}{16}-\frac{3}{2}b+b^{2}=\frac{1267}{16}$$

Subtract \frac{1267}{16} from both sides.

$$\frac{265}{16}-\frac{3}{2}b+b^{2}-\frac{1267}{16}=0$$

Subtract \frac{1267}{16} from \frac{265}{16} to get -\frac{501}{8}.

$$-\frac{501}{8}-\frac{3}{2}b+b^{2}=0$$

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.

$$b^{2}-\frac{3}{2}b-\frac{501}{8}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{3}{2} for b, and -\frac{501}{8} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\left(-\frac{3}{2}\right)^{2}-4\left(-\frac{501}{8}\right)}}{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$b=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}-4\left(-\frac{501}{8}\right)}}{2}$$

Multiply -4 times -\frac{501}{8}.

$$b=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{9}{4}+\frac{501}{2}}}{2}$$

Add \frac{9}{4} to \frac{501}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b=\frac{-\left(-\frac{3}{2}\right)±\sqrt{\frac{1011}{4}}}{2}$$

Take the square root of \frac{1011}{4}.

$$b=\frac{-\left(-\frac{3}{2}\right)±\frac{\sqrt{1011}}{2}}{2}$$

The opposite of -\frac{3}{2} is \frac{3}{2}.

$$b=\frac{\frac{3}{2}±\frac{\sqrt{1011}}{2}}{2}$$

Now solve the equation b=\frac{\frac{3}{2}±\frac{\sqrt{1011}}{2}}{2} when ± is plus. Add \frac{3}{2} to \frac{\sqrt{1011}}{2}.

$$b=\frac{\sqrt{1011}+3}{2\times 2}$$

Divide \frac{3+\sqrt{1011}}{2} by 2.

$$b=\frac{\sqrt{1011}+3}{4}$$

Now solve the equation b=\frac{\frac{3}{2}±\frac{\sqrt{1011}}{2}}{2} when ± is minus. Subtract \frac{\sqrt{1011}}{2} from \frac{3}{2}.

$$b=\frac{3-\sqrt{1011}}{2\times 2}$$

Divide \frac{3-\sqrt{1011}}{2} by 2.

$$b=\frac{3-\sqrt{1011}}{4}$$

The equation is now solved.

$$b=\frac{\sqrt{1011}+3}{4} b=\frac{3-\sqrt{1011}}{4}$$

Steps for Completing the Square

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$5^{2}+\left(\frac{25\sqrt{3}}{4}-2\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$25+\left(\frac{25\sqrt{3}}{4}-2\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Combine \frac{25\sqrt{3}}{4} and -2\sqrt{3} to get \frac{17}{4}\sqrt{3}.

$$25+\left(\frac{17}{4}\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Expand \left(\frac{17}{4}\sqrt{3}\right)^{2}.

$$25+\left(\frac{17}{4}\right)^{2}\left(\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Calculate \frac{17}{4} to the power of 2 and get \frac{289}{16}.

$$25+\frac{289}{16}\left(\sqrt{3}\right)^{2}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

The square of \sqrt{3} is 3.

$$25+\frac{289}{16}\times 3=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Multiply \frac{289}{16} and 3 to get \frac{867}{16}.

$$25+\frac{867}{16}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Add 25 and \frac{867}{16} to get \frac{1267}{16}.

$$\frac{1267}{16}=4^{2}+\left(\frac{3}{4}-b\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{1267}{16}=16+\left(\frac{3}{4}-b\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{4}-b\right)^{2}.

$$\frac{1267}{16}=16+\frac{9}{16}-\frac{3}{2}b+b^{2}$$

Add 16 and \frac{9}{16} to get \frac{265}{16}.

$$\frac{1267}{16}=\frac{265}{16}-\frac{3}{2}b+b^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{265}{16}-\frac{3}{2}b+b^{2}=\frac{1267}{16}$$

Subtract \frac{265}{16} from both sides.

$$-\frac{3}{2}b+b^{2}=\frac{1267}{16}-\frac{265}{16}$$

Subtract \frac{265}{16} from \frac{1267}{16} to get \frac{501}{8}.

$$-\frac{3}{2}b+b^{2}=\frac{501}{8}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$b^{2}-\frac{3}{2}b=\frac{501}{8}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-\frac{3}{2}b+\left(-\frac{3}{4}\right)^{2}=\frac{501}{8}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$b^{2}-\frac{3}{2}b+\frac{9}{16}=\frac{501}{8}+\frac{9}{16}$$

Add \frac{501}{8} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b^{2}-\frac{3}{2}b+\frac{9}{16}=\frac{1011}{16}$$

Factor b^{2}-\frac{3}{2}b+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-\frac{3}{4}\right)^{2}=\frac{1011}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-\frac{3}{4}\right)^{2}}=\sqrt{\frac{1011}{16}}$$

Simplify.

$$b-\frac{3}{4}=\frac{\sqrt{1011}}{4} b-\frac{3}{4}=-\frac{\sqrt{1011}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$b=\frac{\sqrt{1011}+3}{4} b=\frac{3-\sqrt{1011}}{4}$$