Solve (sqrt(5) - 2)/(sqrt(5 + 2)) - (sqrt(5) + 2)/(sqrt(5) - 2) | 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

$$\frac{\sqrt{5}-2}{\sqrt{5+2}}-\frac{\sqrt{5}+2}{\sqrt{5}-2}$$

Evaluate

$\frac{\sqrt{35}}{7}-\frac{2\sqrt{7}}{7}-4\sqrt{5}-9\approx -17.855046601$

Short Solution Steps

Add $5$ and $2$ to get $7$.

$$\frac{\sqrt{5}-2}{\sqrt{7}}-\frac{\sqrt{5}+2}{\sqrt{5}-2}$$

Rationalize the denominator of $\frac{\sqrt{5}-2}{\sqrt{7}}$ by multiplying numerator and denominator by $\sqrt{7}$.

$$\frac{\left(\sqrt{5}-2\right)\sqrt{7}}{\left(\sqrt{7}\right)^{2}}-\frac{\sqrt{5}+2}{\sqrt{5}-2}$$

The square of $\sqrt{7}$ is $7$.

$$\frac{\left(\sqrt{5}-2\right)\sqrt{7}}{7}-\frac{\sqrt{5}+2}{\sqrt{5}-2}$$

Rationalize the denominator of $\frac{\sqrt{5}+2}{\sqrt{5}-2}$ by multiplying numerator and denominator by $\sqrt{5}+2$.

$$\frac{\left(\sqrt{5}-2\right)\sqrt{7}}{7}-\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}$$

Consider $\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(\sqrt{5}-2\right)\sqrt{7}}{7}-\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}\right)^{2}-2^{2}}$$

Square $\sqrt{5}$. Square $2$.

$$\frac{\left(\sqrt{5}-2\right)\sqrt{7}}{7}-\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}+2\right)}{5-4}$$

Subtract $4$ from $5$ to get $1$.

$$\frac{\left(\sqrt{5}-2\right)\sqrt{7}}{7}-\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}+2\right)}{1}$$

Anything divided by one gives itself.

$$\frac{\left(\sqrt{5}-2\right)\sqrt{7}}{7}-\left(\sqrt{5}+2\right)\left(\sqrt{5}+2\right)$$

Multiply $\sqrt{5}+2$ and $\sqrt{5}+2$ to get $\left(\sqrt{5}+2\right)^{2}$.

$$\frac{\left(\sqrt{5}-2\right)\sqrt{7}}{7}-\left(\sqrt{5}+2\right)^{2}$$

Use the distributive property to multiply $\sqrt{5}-2$ by $\sqrt{7}$.

$$\frac{\sqrt{5}\sqrt{7}-2\sqrt{7}}{7}-\left(\sqrt{5}+2\right)^{2}$$

To multiply $\sqrt{5}$ and $\sqrt{7}$, multiply the numbers under the square root.

$$\frac{\sqrt{35}-2\sqrt{7}}{7}-\left(\sqrt{5}+2\right)^{2}$$

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(\sqrt{5}+2\right)^{2}$.

$$\frac{\sqrt{35}-2\sqrt{7}}{7}-\left(\left(\sqrt{5}\right)^{2}+4\sqrt{5}+4\right)$$

The square of $\sqrt{5}$ is $5$.

$$\frac{\sqrt{35}-2\sqrt{7}}{7}-\left(5+4\sqrt{5}+4\right)$$

Add $5$ and $4$ to get $9$.

$$\frac{\sqrt{35}-2\sqrt{7}}{7}-\left(9+4\sqrt{5}\right)$$

To find the opposite of $9+4\sqrt{5}$, find the opposite of each term.

$$\frac{\sqrt{35}-2\sqrt{7}}{7}-9-4\sqrt{5}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $-9-4\sqrt{5}$ times $\frac{7}{7}$.

$$\frac{\sqrt{35}-2\sqrt{7}}{7}+\frac{7\left(-9-4\sqrt{5}\right)}{7}$$

Since $\frac{\sqrt{35}-2\sqrt{7}}{7}$ and $\frac{7\left(-9-4\sqrt{5}\right)}{7}$ have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{35}-2\sqrt{7}+7\left(-9-4\sqrt{5}\right)}{7}$$

Do the multiplications in $\sqrt{35}-2\sqrt{7}+7\left(-9-4\sqrt{5}\right)$.

$$\frac{\sqrt{35}-2\sqrt{7}-63-28\sqrt{5}}{7}$$

Factor

$\frac{\sqrt{35} - 2 \sqrt{7} - 28 \sqrt{5} - 63}{7} = -17.855046601289096$