Solve (5-1/2x)^2+(frac{sqrt{3}}{2}x)^2=(14/3sqrt{3})^2

$$( 5 - \frac { 1 } { 2 } x ) ^ { 2 } + ( \frac { \sqrt { 3 } } { 2 } x ) ^ { 2 } = ( \frac { 14 } { 3 } \sqrt { 3 } ) ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{1677}}{6}+\frac{5}{2}\approx 9.32519841$$

$$x=-\frac{\sqrt{1677}}{6}+\frac{5}{2}\approx -4.32519841$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\frac{1}{2}x\right)^{2}.

$$25-5x+\frac{1}{4}x^{2}+\left(\frac{\sqrt{3}}{2}x\right)^{2}=\left(\frac{14}{3}\sqrt{3}\right)^{2}$$

Express \frac{\sqrt{3}}{2}x as a single fraction.

$$25-5x+\frac{1}{4}x^{2}+\left(\frac{\sqrt{3}x}{2}\right)^{2}=\left(\frac{14}{3}\sqrt{3}\right)^{2}$$

To raise \frac{\sqrt{3}x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\left(\frac{14}{3}\sqrt{3}\right)^{2}$$

Expand \left(\frac{14}{3}\sqrt{3}\right)^{2}.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\left(\frac{14}{3}\right)^{2}\left(\sqrt{3}\right)^{2}$$

Calculate \frac{14}{3} to the power of 2 and get \frac{196}{9}.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\frac{196}{9}\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\frac{196}{9}\times 3$$

Multiply \frac{196}{9} and 3 to get \frac{196}{3}.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\frac{196}{3}$$

Expand \left(\sqrt{3}x\right)^{2}.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}\right)^{2}x^{2}}{2^{2}}=\frac{196}{3}$$

The square of \sqrt{3} is 3.

$$25-5x+\frac{1}{4}x^{2}+\frac{3x^{2}}{2^{2}}=\frac{196}{3}$$

Calculate 2 to the power of 2 and get 4.

$$25-5x+\frac{1}{4}x^{2}+\frac{3x^{2}}{4}=\frac{196}{3}$$

Subtract \frac{196}{3} from both sides.

$$25-5x+\frac{1}{4}x^{2}+\frac{3x^{2}}{4}-\frac{196}{3}=0$$

Subtract \frac{196}{3} from 25 to get -\frac{121}{3}.

$$-\frac{121}{3}-5x+\frac{1}{4}x^{2}+\frac{3x^{2}}{4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 4 is 12. Multiply -\frac{121}{3} times \frac{4}{4}. Multiply \frac{3x^{2}}{4} times \frac{3}{3}.

$$-\frac{121\times 4}{12}-5x+\frac{1}{4}x^{2}+\frac{3\times 3x^{2}}{12}=0$$

Since -\frac{121\times 4}{12} and \frac{3\times 3x^{2}}{12} have the same denominator, add them by adding their numerators.

$$\frac{-121\times 4+3\times 3x^{2}}{12}-5x+\frac{1}{4}x^{2}=0$$

Do the multiplications in -121\times 4+3\times 3x^{2}.

$$\frac{-484+9x^{2}}{12}-5x+\frac{1}{4}x^{2}=0$$

Multiply both sides of the equation by 12, the least common multiple of 12,4.

$$-484+9x^{2}-60x+3x^{2}=0$$

Reorder the terms.

$$3x^{2}+9x^{2}-60x-484=0$$

Combine 3x^{2} and 9x^{2} to get 12x^{2}.

$$12x^{2}-60x-484=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -60 for b, and -484 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-60\right)±\sqrt{\left(-60\right)^{2}-4\times 12\left(-484\right)}}{2\times 12}$$

Square -60.

$$x=\frac{-\left(-60\right)±\sqrt{3600-4\times 12\left(-484\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-60\right)±\sqrt{3600-48\left(-484\right)}}{2\times 12}$$

Multiply -48 times -484.

$$x=\frac{-\left(-60\right)±\sqrt{3600+23232}}{2\times 12}$$

Add 3600 to 23232.

$$x=\frac{-\left(-60\right)±\sqrt{26832}}{2\times 12}$$

Take the square root of 26832.

$$x=\frac{-\left(-60\right)±4\sqrt{1677}}{2\times 12}$$

The opposite of -60 is 60.

$$x=\frac{60±4\sqrt{1677}}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{60±4\sqrt{1677}}{24}$$

Now solve the equation x=\frac{60±4\sqrt{1677}}{24} when ± is plus. Add 60 to 4\sqrt{1677}.

$$x=\frac{4\sqrt{1677}+60}{24}$$

Divide 60+4\sqrt{1677} by 24.

$$x=\frac{\sqrt{1677}}{6}+\frac{5}{2}$$

Now solve the equation x=\frac{60±4\sqrt{1677}}{24} when ± is minus. Subtract 4\sqrt{1677} from 60.

$$x=\frac{60-4\sqrt{1677}}{24}$$

Divide 60-4\sqrt{1677} by 24.

$$x=-\frac{\sqrt{1677}}{6}+\frac{5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{1677}}{6}+\frac{5}{2} x=-\frac{\sqrt{1677}}{6}+\frac{5}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\frac{1}{2}x\right)^{2}.

$$25-5x+\frac{1}{4}x^{2}+\left(\frac{\sqrt{3}}{2}x\right)^{2}=\left(\frac{14}{3}\sqrt{3}\right)^{2}$$

Express \frac{\sqrt{3}}{2}x as a single fraction.

$$25-5x+\frac{1}{4}x^{2}+\left(\frac{\sqrt{3}x}{2}\right)^{2}=\left(\frac{14}{3}\sqrt{3}\right)^{2}$$

To raise \frac{\sqrt{3}x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\left(\frac{14}{3}\sqrt{3}\right)^{2}$$

Expand \left(\frac{14}{3}\sqrt{3}\right)^{2}.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\left(\frac{14}{3}\right)^{2}\left(\sqrt{3}\right)^{2}$$

Calculate \frac{14}{3} to the power of 2 and get \frac{196}{9}.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\frac{196}{9}\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\frac{196}{9}\times 3$$

Multiply \frac{196}{9} and 3 to get \frac{196}{3}.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}x\right)^{2}}{2^{2}}=\frac{196}{3}$$

Expand \left(\sqrt{3}x\right)^{2}.

$$25-5x+\frac{1}{4}x^{2}+\frac{\left(\sqrt{3}\right)^{2}x^{2}}{2^{2}}=\frac{196}{3}$$

The square of \sqrt{3} is 3.

$$25-5x+\frac{1}{4}x^{2}+\frac{3x^{2}}{2^{2}}=\frac{196}{3}$$

Calculate 2 to the power of 2 and get 4.

$$25-5x+\frac{1}{4}x^{2}+\frac{3x^{2}}{4}=\frac{196}{3}$$

Subtract 25 from both sides.

$$-5x+\frac{1}{4}x^{2}+\frac{3x^{2}}{4}=\frac{196}{3}-25$$

Subtract 25 from \frac{196}{3} to get \frac{121}{3}.

$$-5x+\frac{1}{4}x^{2}+\frac{3x^{2}}{4}=\frac{121}{3}$$

Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$-60x+3x^{2}+3\times 3x^{2}=484$$

Reorder the terms.

$$3\times 3x^{2}+3x^{2}-60x=484$$

Multiply 3 and 3 to get 9.

$$9x^{2}+3x^{2}-60x=484$$

Combine 9x^{2} and 3x^{2} to get 12x^{2}.

$$12x^{2}-60x=484$$

Divide both sides by 12.

$$\frac{12x^{2}-60x}{12}=\frac{484}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{60}{12}\right)x=\frac{484}{12}$$

Divide -60 by 12.

$$x^{2}-5x=\frac{484}{12}$$

Reduce the fraction \frac{484}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}-5x=\frac{121}{3}$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=\frac{121}{3}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=\frac{121}{3}+\frac{25}{4}$$

Add \frac{121}{3} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=\frac{559}{12}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{559}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{559}{12}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{1677}}{6} x-\frac{5}{2}=-\frac{\sqrt{1677}}{6}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{1677}}{6}+\frac{5}{2} x=-\frac{\sqrt{1677}}{6}+\frac{5}{2}$$