Solve I = (sqrt(2) * x - sqrt(3)) ^ 2 - (sqrt(3) * x + 2sqrt(2)) ^ 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

$$I=(\sqrt{2}x-\sqrt{3})^{2}-(\sqrt{3}x+2\sqrt{2})^{2}$$

Solve for I

$I=-\left(x^{2}+6\sqrt{6}x+5\right)$

Solution Steps

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

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

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

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

To multiply $\sqrt{2}$ and $\sqrt{3}$, multiply the numbers under the square root.

$$I=2x^{2}-2\sqrt{6}x+\left(\sqrt{3}\right)^{2}-\left(\sqrt{3}x+2\sqrt{2}\right)^{2}$$

The square of $\sqrt{3}$ is $3$.

$$I=2x^{2}-2\sqrt{6}x+3-\left(\sqrt{3}x+2\sqrt{2}\right)^{2}$$

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

$$I=2x^{2}-2\sqrt{6}x+3-\left(\left(\sqrt{3}\right)^{2}x^{2}+4\sqrt{3}x\sqrt{2}+4\left(\sqrt{2}\right)^{2}\right)$$

The square of $\sqrt{3}$ is $3$.

$$I=2x^{2}-2\sqrt{6}x+3-\left(3x^{2}+4\sqrt{3}x\sqrt{2}+4\left(\sqrt{2}\right)^{2}\right)$$

To multiply $\sqrt{3}$ and $\sqrt{2}$, multiply the numbers under the square root.

$$I=2x^{2}-2\sqrt{6}x+3-\left(3x^{2}+4\sqrt{6}x+4\left(\sqrt{2}\right)^{2}\right)$$

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

$$I=2x^{2}-2\sqrt{6}x+3-\left(3x^{2}+4\sqrt{6}x+4\times 2\right)$$

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

$$I=2x^{2}-2\sqrt{6}x+3-\left(3x^{2}+4\sqrt{6}x+8\right)$$

To find the opposite of $3x^{2}+4\sqrt{6}x+8$, find the opposite of each term.

$$I=2x^{2}-2\sqrt{6}x+3-3x^{2}-4\sqrt{6}x-8$$

Combine $2x^{2}$ and $-3x^{2}$ to get $-x^{2}$.

$$I=-x^{2}-2\sqrt{6}x+3-4\sqrt{6}x-8$$

Combine $-2\sqrt{6}x$ and $-4\sqrt{6}x$ to get $-6\sqrt{6}x$.

$$I=-x^{2}-6\sqrt{6}x+3-8$$

Subtract $8$ from $3$ to get $-5$.

$$I=-x^{2}-6\sqrt{6}x-5$$

Solve for x (complex solution)

$x=\sqrt{49-I}-3\sqrt{6}$
$x=-\sqrt{49-I}-3\sqrt{6}$

Solve for x

$x=\sqrt{49-I}-3\sqrt{6}$
$x=-\sqrt{49-I}-3\sqrt{6}\text{, }I\leq 49$