Solve 2{left(frac{3+sqrt{33}}{4}right)}^2-3(frac{3+sqrt{33}}{4})-3=0

$$2 { \left( \frac{ 3+ \sqrt{ 33 } }{ 4 } \right) }^{ 2 } -3( \frac{ 3+ \sqrt{ 33 } }{ 4 } )-3 = 0$$

Verify

Solution Steps

Multiply both sides of the equation by 4.

$$4\left(2\times \left(\frac{3+\sqrt{33}}{4}\right)^{2}-3\times \frac{3+\sqrt{33}}{4}\right)-12=0$$

To raise \frac{3+\sqrt{33}}{4} to a power, raise both numerator and denominator to the power and then divide.

$$4\left(2\times \frac{\left(3+\sqrt{33}\right)^{2}}{4^{2}}-3\times \frac{3+\sqrt{33}}{4}\right)-12=0$$

Express 2\times \frac{\left(3+\sqrt{33}\right)^{2}}{4^{2}} as a single fraction.

$$4\left(\frac{2\left(3+\sqrt{33}\right)^{2}}{4^{2}}-3\times \frac{3+\sqrt{33}}{4}\right)-12=0$$

Express 3\times \frac{3+\sqrt{33}}{4} as a single fraction.

$$4\left(\frac{2\left(3+\sqrt{33}\right)^{2}}{4^{2}}-\frac{3\left(3+\sqrt{33}\right)}{4}\right)-12=0$$

Use the distributive property to multiply 3 by 3+\sqrt{33}.

$$4\left(\frac{2\left(3+\sqrt{33}\right)^{2}}{4^{2}}-\frac{9+3\sqrt{33}}{4}\right)-12=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4^{2} and 4 is 16. Multiply \frac{9+3\sqrt{33}}{4} times \frac{4}{4}.

$$4\left(\frac{2\left(3+\sqrt{33}\right)^{2}}{16}-\frac{4\left(9+3\sqrt{33}\right)}{16}\right)-12=0$$

Since \frac{2\left(3+\sqrt{33}\right)^{2}}{16} and \frac{4\left(9+3\sqrt{33}\right)}{16} have the same denominator, subtract them by subtracting their numerators.

$$4\times \frac{2\left(3+\sqrt{33}\right)^{2}-4\left(9+3\sqrt{33}\right)}{16}-12=0$$

Cancel out 16, the greatest common factor in 4 and 16.

$$\frac{2\left(3+\sqrt{33}\right)^{2}-4\left(9+3\sqrt{33}\right)}{4}-12=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 12 times \frac{4}{4}.

$$\frac{2\left(3+\sqrt{33}\right)^{2}-4\left(9+3\sqrt{33}\right)}{4}-\frac{12\times 4}{4}=0$$

Since \frac{2\left(3+\sqrt{33}\right)^{2}-4\left(9+3\sqrt{33}\right)}{4} and \frac{12\times 4}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(3+\sqrt{33}\right)^{2}-4\left(9+3\sqrt{33}\right)-12\times 4}{4}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{33}\right)^{2}.

$$\frac{2\left(9+6\sqrt{33}+\left(\sqrt{33}\right)^{2}\right)-4\left(9+3\sqrt{33}\right)-12\times 4}{4}=0$$

The square of \sqrt{33} is 33.

$$\frac{2\left(9+6\sqrt{33}+33\right)-4\left(9+3\sqrt{33}\right)-12\times 4}{4}=0$$

Add 9 and 33 to get 42.

$$\frac{2\left(42+6\sqrt{33}\right)-4\left(9+3\sqrt{33}\right)-12\times 4}{4}=0$$

Multiply -12 and 4 to get -48.

$$\frac{2\left(42+6\sqrt{33}\right)-4\left(9+3\sqrt{33}\right)-48}{4}=0$$

Multiply both sides of the equation by 4.

$$2\left(42+6\sqrt{33}\right)-4\left(9+3\sqrt{33}\right)-48=0$$

Use the distributive property to multiply 2 by 42+6\sqrt{33}.

$$84+12\sqrt{33}-4\left(9+3\sqrt{33}\right)-48=0$$

Use the distributive property to multiply -4 by 9+3\sqrt{33}.

$$84+12\sqrt{33}-36-12\sqrt{33}-48=0$$

Subtract 36 from 84 to get 48.

$$48+12\sqrt{33}-12\sqrt{33}-48=0$$

Combine 12\sqrt{33} and -12\sqrt{33} to get 0.

$$48-48=0$$

Subtract 48 from 48 to get 0.

$$0=0$$

Compare 0 and 0.

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