Solve (sqrt(5) + sqrt(2)) ^ 2 + (sqrt(8) - 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

$$(\sqrt{5}+\sqrt{2})^{2}+(\sqrt{8}-\sqrt{5})^{2}$$

Evaluate

$20-2\sqrt{10}\approx 13.67544468$

Solution Steps

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

$$\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

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

$$5+2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

To multiply $\sqrt{5}$ and $\sqrt{2}$, multiply the numbers under the square root.

$$5+2\sqrt{10}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

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

$$5+2\sqrt{10}+2+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

Factor $8=2^{2}\times 2$. Rewrite the square root of the product $\sqrt{2^{2}\times 2}$ as the product of square roots $\sqrt{2^{2}}\sqrt{2}$. Take the square root of $2^{2}$.

$$7+2\sqrt{10}+\left(2\sqrt{2}-\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+4\left(\sqrt{2}\right)^{2}-4\sqrt{2}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+4\times 2-4\sqrt{2}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+8-4\sqrt{2}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

To multiply $\sqrt{2}$ and $\sqrt{5}$, multiply the numbers under the square root.

$$7+2\sqrt{10}+8-4\sqrt{10}+\left(\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+8-4\sqrt{10}+5$$

Add $8$ and $5$ to get $13$.

$$7+2\sqrt{10}+13-4\sqrt{10}$$

Add $7$ and $13$ to get $20$.

$$20+2\sqrt{10}-4\sqrt{10}$$

Combine $2\sqrt{10}$ and $-4\sqrt{10}$ to get $-2\sqrt{10}$.

$$20-2\sqrt{10}$$

Expand

$20-2\sqrt{10}$

Solution Steps

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

$$\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

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

$$5+2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

To multiply $\sqrt{5}$ and $\sqrt{2}$, multiply the numbers under the square root.

$$5+2\sqrt{10}+\left(\sqrt{2}\right)^{2}+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

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

$$5+2\sqrt{10}+2+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+\left(\sqrt{8}-\sqrt{5}\right)^{2}$$

Factor $8=2^{2}\times 2$. Rewrite the square root of the product $\sqrt{2^{2}\times 2}$ as the product of square roots $\sqrt{2^{2}}\sqrt{2}$. Take the square root of $2^{2}$.

$$7+2\sqrt{10}+\left(2\sqrt{2}-\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+4\left(\sqrt{2}\right)^{2}-4\sqrt{2}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+4\times 2-4\sqrt{2}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+8-4\sqrt{2}\sqrt{5}+\left(\sqrt{5}\right)^{2}$$

To multiply $\sqrt{2}$ and $\sqrt{5}$, multiply the numbers under the square root.

$$7+2\sqrt{10}+8-4\sqrt{10}+\left(\sqrt{5}\right)^{2}$$

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

$$7+2\sqrt{10}+8-4\sqrt{10}+5$$

Add $8$ and $5$ to get $13$.

$$7+2\sqrt{10}+13-4\sqrt{10}$$

Add $7$ and $13$ to get $20$.

$$20+2\sqrt{10}-4\sqrt{10}$$

Combine $2\sqrt{10}$ and $-4\sqrt{10}$ to get $-2\sqrt{10}$.

$$20-2\sqrt{10}$$