Solve H = \frac { 1 } { ( 1 - \sqrt { 2 } ) ^ { 2 } }

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Question

$$H = \frac { 1 } { ( 1 - \sqrt { 2 } ) ^ { 2 } }$$

Solve for H

$H=2\sqrt{2}+3\approx 5.828427125$

Steps for Solving Linear Equation

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

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

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

$$H=\frac{1}{1-2\sqrt{2}+2}$$

Add $1$ and $2$ to get $3$.

$$H=\frac{1}{3-2\sqrt{2}}$$

Rationalize the denominator of $\frac{1}{3-2\sqrt{2}}$ by multiplying numerator and denominator by $3+2\sqrt{2}$.

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

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

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

Calculate $3$ to the power of $2$ and get $9$.

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

Expand $\left(-2\sqrt{2}\right)^{2}$.

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

Calculate $-2$ to the power of $2$ and get $4$.

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

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

$$H=\frac{3+2\sqrt{2}}{9-4\times 2}$$

Multiply $4$ and $2$ to get $8$.

$$H=\frac{3+2\sqrt{2}}{9-8}$$

Subtract $8$ from $9$ to get $1$.

$$H=\frac{3+2\sqrt{2}}{1}$$

Anything divided by one gives itself.

$$H=3+2\sqrt{2}$$

Show Solution

Assign H

$H≔2\sqrt{2}+3$