Solve If x = (3 + sqrt(2))/(3 - sqrt(2)) then find x + 1/x | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x = \frac { 3 + \sqrt { 2 } } { 3 - \sqrt { 2 } } x + \frac { 1 } { x }$$

Solve for x (complex solution)

$x=-\frac{i\sqrt{3\sqrt{2}-2}}{2}\approx -0-0.748772443i$
$x=\frac{i\sqrt{3\sqrt{2}-2}}{2}\approx 0.748772443i$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Rationalize the denominator of $\frac{3+\sqrt{2}}{3-\sqrt{2}}$ by multiplying numerator and denominator by $3+\sqrt{2}$.

$$x=\frac{\left(3+\sqrt{2}\right)\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}x+\frac{1}{x}$$

Consider $\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

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

Square $3$. Square $\sqrt{2}$.

$$x=\frac{\left(3+\sqrt{2}\right)\left(3+\sqrt{2}\right)}{9-2}x+\frac{1}{x}$$

Subtract $2$ from $9$ to get $7$.

$$x=\frac{\left(3+\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}x+\frac{1}{x}$$

Multiply $3+\sqrt{2}$ and $3+\sqrt{2}$ to get $\left(3+\sqrt{2}\right)^{2}$.

$$x=\frac{\left(3+\sqrt{2}\right)^{2}}{7}x+\frac{1}{x}$$

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

$$x=\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7}x+\frac{1}{x}$$

The square of $\sqrt{2}$ is $2$.

$$x=\frac{9+6\sqrt{2}+2}{7}x+\frac{1}{x}$$

Add $9$ and $2$ to get $11$.

$$x=\frac{11+6\sqrt{2}}{7}x+\frac{1}{x}$$

Express $\frac{11+6\sqrt{2}}{7}x$ as a single fraction.

$$x=\frac{\left(11+6\sqrt{2}\right)x}{7}+\frac{1}{x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $7$ and $x$ is $7x$. Multiply $\frac{\left(11+6\sqrt{2}\right)x}{7}$ times $\frac{x}{x}$. Multiply $\frac{1}{x}$ times $\frac{7}{7}$.

$$x=\frac{\left(11+6\sqrt{2}\right)xx}{7x}+\frac{7}{7x}$$

Since $\frac{\left(11+6\sqrt{2}\right)xx}{7x}$ and $\frac{7}{7x}$ have the same denominator, add them by adding their numerators.

$$x=\frac{\left(11+6\sqrt{2}\right)xx+7}{7x}$$

Do the multiplications in $\left(11+6\sqrt{2}\right)xx+7$.

$$x=\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}$$

Subtract $\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}$ from both sides.

$$x-\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $x$ times $\frac{7x}{7x}$.

$$\frac{x\times 7x}{7x}-\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}=0$$

Since $\frac{x\times 7x}{7x}$ and $\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 7x-\left(11x^{2}+6\sqrt{2}x^{2}+7\right)}{7x}=0$$

Do the multiplications in $x\times 7x-\left(11x^{2}+6\sqrt{2}x^{2}+7\right)$.

$$\frac{7x^{2}-11x^{2}-6\sqrt{2}x^{2}-7}{7x}=0$$

Combine like terms in $7x^{2}-11x^{2}-6\sqrt{2}x^{2}-7$.

$$\frac{-6\sqrt{2}x^{2}-4x^{2}-7}{7x}=0$$

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $7x$.

$$-6\sqrt{2}x^{2}-4x^{2}-7=0$$

Add $7$ to both sides. Anything plus zero gives itself.

$$-6\sqrt{2}x^{2}-4x^{2}=7$$

Combine all terms containing $x$.

$$\left(-6\sqrt{2}-4\right)x^{2}=7$$

Dividing by $-6\sqrt{2}-4$ undoes the multiplication by $-6\sqrt{2}-4$.

$$x^{2}=\frac{7}{-6\sqrt{2}-4}$$

Divide $7$ by $-6\sqrt{2}-4$.

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

Take the square root of both sides of the equation.

$$x=\frac{i\sqrt{3-\sqrt{2}}}{2^{\frac{3}{4}}}$$ $$x=-\frac{i\sqrt{3-\sqrt{2}}}{2^{\frac{3}{4}}}$$

Steps Using the Quadratic Formula

Rationalize the denominator of $\frac{3+\sqrt{2}}{3-\sqrt{2}}$ by multiplying numerator and denominator by $3+\sqrt{2}$.

$$x=\frac{\left(3+\sqrt{2}\right)\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}x+\frac{1}{x}$$

Consider $\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

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

Square $3$. Square $\sqrt{2}$.

$$x=\frac{\left(3+\sqrt{2}\right)\left(3+\sqrt{2}\right)}{9-2}x+\frac{1}{x}$$

Subtract $2$ from $9$ to get $7$.

$$x=\frac{\left(3+\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}x+\frac{1}{x}$$

Multiply $3+\sqrt{2}$ and $3+\sqrt{2}$ to get $\left(3+\sqrt{2}\right)^{2}$.

$$x=\frac{\left(3+\sqrt{2}\right)^{2}}{7}x+\frac{1}{x}$$

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

$$x=\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7}x+\frac{1}{x}$$

The square of $\sqrt{2}$ is $2$.

$$x=\frac{9+6\sqrt{2}+2}{7}x+\frac{1}{x}$$

Add $9$ and $2$ to get $11$.

$$x=\frac{11+6\sqrt{2}}{7}x+\frac{1}{x}$$

Express $\frac{11+6\sqrt{2}}{7}x$ as a single fraction.

$$x=\frac{\left(11+6\sqrt{2}\right)x}{7}+\frac{1}{x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $7$ and $x$ is $7x$. Multiply $\frac{\left(11+6\sqrt{2}\right)x}{7}$ times $\frac{x}{x}$. Multiply $\frac{1}{x}$ times $\frac{7}{7}$.

$$x=\frac{\left(11+6\sqrt{2}\right)xx}{7x}+\frac{7}{7x}$$

Since $\frac{\left(11+6\sqrt{2}\right)xx}{7x}$ and $\frac{7}{7x}$ have the same denominator, add them by adding their numerators.

$$x=\frac{\left(11+6\sqrt{2}\right)xx+7}{7x}$$

Do the multiplications in $\left(11+6\sqrt{2}\right)xx+7$.

$$x=\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}$$

Subtract $\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}$ from both sides.

$$x-\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $x$ times $\frac{7x}{7x}$.

$$\frac{x\times 7x}{7x}-\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}=0$$

Since $\frac{x\times 7x}{7x}$ and $\frac{11x^{2}+6\sqrt{2}x^{2}+7}{7x}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\times 7x-\left(11x^{2}+6\sqrt{2}x^{2}+7\right)}{7x}=0$$

Do the multiplications in $x\times 7x-\left(11x^{2}+6\sqrt{2}x^{2}+7\right)$.

$$\frac{7x^{2}-11x^{2}-6\sqrt{2}x^{2}-7}{7x}=0$$

Combine like terms in $7x^{2}-11x^{2}-6\sqrt{2}x^{2}-7$.

$$\frac{-6\sqrt{2}x^{2}-4x^{2}-7}{7x}=0$$

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $7x$.

$$-6\sqrt{2}x^{2}-4x^{2}-7=0$$

Combine all terms containing $x$.

$$\left(-6\sqrt{2}-4\right)x^{2}-7=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-6\sqrt{2}-4$ for $a$, $0$ for $b$, and $-7$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\left(-6\sqrt{2}-4\right)\left(-7\right)}}{2\left(-6\sqrt{2}-4\right)}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\left(-6\sqrt{2}-4\right)\left(-7\right)}}{2\left(-6\sqrt{2}-4\right)}$$

Multiply $-4$ times $-6\sqrt{2}-4$.

$$x=\frac{0±\sqrt{\left(24\sqrt{2}+16\right)\left(-7\right)}}{2\left(-6\sqrt{2}-4\right)}$$

Multiply $24\sqrt{2}+16$ times $-7$.

$$x=\frac{0±\sqrt{-168\sqrt{2}-112}}{2\left(-6\sqrt{2}-4\right)}$$

Take the square root of $-168\sqrt{2}-112$.

$$x=\frac{0±2i\sqrt{42\sqrt{2}+28}}{2\left(-6\sqrt{2}-4\right)}$$

Multiply $2$ times $-6\sqrt{2}-4$.

$$x=\frac{0±2i\sqrt{42\sqrt{2}+28}}{-12\sqrt{2}-8}$$

Now solve the equation $x=\frac{0±2i\sqrt{42\sqrt{2}+28}}{-12\sqrt{2}-8}$ when $±$ is plus.

$$x=-\frac{\sqrt[4]{2}i\sqrt{3-\sqrt{2}}}{2}$$

Now solve the equation $x=\frac{0±2i\sqrt{42\sqrt{2}+28}}{-12\sqrt{2}-8}$ when $±$ is minus.

$$x=\frac{\sqrt[4]{2}i\sqrt{3-\sqrt{2}}}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt[4]{2}i\sqrt{3-\sqrt{2}}}{2}$$ $$x=\frac{\sqrt[4]{2}i\sqrt{3-\sqrt{2}}}{2}$$