1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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Question

$$( 2 \sqrt { 3 } + 5 \sqrt { 2 } ) \div ( 7 \sqrt { 2 } - 4 \sqrt { 3 } )$$

Evaluate

$\frac{17\sqrt{6}+47}{25}\approx 3.545653025$

Solution Steps

Rationalize the denominator of $\frac{2\sqrt{3}+5\sqrt{2}}{7\sqrt{2}-4\sqrt{3}}$ by multiplying numerator and denominator by $7\sqrt{2}+4\sqrt{3}$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{\left(7\sqrt{2}-4\sqrt{3}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}$$

Consider $\left(7\sqrt{2}-4\sqrt{3}\right)\left(7\sqrt{2}+4\sqrt{3}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{\left(7\sqrt{2}\right)^{2}-\left(-4\sqrt{3}\right)^{2}}$$

Expand $\left(7\sqrt{2}\right)^{2}$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{7^{2}\left(\sqrt{2}\right)^{2}-\left(-4\sqrt{3}\right)^{2}}$$

Calculate $7$ to the power of $2$ and get $49$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{49\left(\sqrt{2}\right)^{2}-\left(-4\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{49\times 2-\left(-4\sqrt{3}\right)^{2}}$$

Multiply $49$ and $2$ to get $98$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{98-\left(-4\sqrt{3}\right)^{2}}$$

Expand $\left(-4\sqrt{3}\right)^{2}$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{98-\left(-4\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate $-4$ to the power of $2$ and get $16$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{98-16\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{98-16\times 3}$$

Multiply $16$ and $3$ to get $48$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{98-48}$$

Subtract $48$ from $98$ to get $50$.

$$\frac{\left(2\sqrt{3}+5\sqrt{2}\right)\left(7\sqrt{2}+4\sqrt{3}\right)}{50}$$

Apply the distributive property by multiplying each term of $2\sqrt{3}+5\sqrt{2}$ by each term of $7\sqrt{2}+4\sqrt{3}$.

$$\frac{14\sqrt{3}\sqrt{2}+8\left(\sqrt{3}\right)^{2}+35\left(\sqrt{2}\right)^{2}+20\sqrt{3}\sqrt{2}}{50}$$

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

$$\frac{14\sqrt{6}+8\left(\sqrt{3}\right)^{2}+35\left(\sqrt{2}\right)^{2}+20\sqrt{3}\sqrt{2}}{50}$$

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

$$\frac{14\sqrt{6}+8\times 3+35\left(\sqrt{2}\right)^{2}+20\sqrt{3}\sqrt{2}}{50}$$

Multiply $8$ and $3$ to get $24$.

$$\frac{14\sqrt{6}+24+35\left(\sqrt{2}\right)^{2}+20\sqrt{3}\sqrt{2}}{50}$$

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

$$\frac{14\sqrt{6}+24+35\times 2+20\sqrt{3}\sqrt{2}}{50}$$

Multiply $35$ and $2$ to get $70$.

$$\frac{14\sqrt{6}+24+70+20\sqrt{3}\sqrt{2}}{50}$$

Add $24$ and $70$ to get $94$.

$$\frac{14\sqrt{6}+94+20\sqrt{3}\sqrt{2}}{50}$$

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

$$\frac{14\sqrt{6}+94+20\sqrt{6}}{50}$$

Combine $14\sqrt{6}$ and $20\sqrt{6}$ to get $34\sqrt{6}$.

$$\frac{34\sqrt{6}+94}{50}$$

Show Solution

Factor

$\frac{17 \sqrt{6} + 47}{25} = 3.5456530250925606$

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