Solve ( 3 + \sqrt { 2 } ) ^ { 2 } | Equatix

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a⋅(b-2)=3b

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Question

$$( 3 + \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$6\sqrt{2}+11\approx 19.485281374$

Solution Steps

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

$$9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

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

$$9+6\sqrt{2}+2$$

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

$$11+6\sqrt{2}$$

Show Solution

Expand

$6 \sqrt{2} + 11 = 19.485281374$

Solution Steps

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

$$9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

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

$$9+6\sqrt{2}+2$$

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

$$11+6\sqrt{2}$$

Show Solution

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Evaluate

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Solution Steps