Solve (2x-(3)/(x))(2x+(3)/(x)) | 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

$$(2x- \frac{ 3 }{ x } )(2x+ \frac{ 3 }{ x } )$$

Evaluate

$4x^{2}-\frac{9}{x^{2}}$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply $2x$ times $\frac{x}{x}$.

$$\left(\frac{2xx}{x}-\frac{3}{x}\right)\left(2x+\frac{3}{x}\right)$$

Since $\frac{2xx}{x}$ and $\frac{3}{x}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{2xx-3}{x}\left(2x+\frac{3}{x}\right)$$

Do the multiplications in $2xx-3$.

$$\frac{2x^{2}-3}{x}\left(2x+\frac{3}{x}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $2x$ times $\frac{x}{x}$.

$$\frac{2x^{2}-3}{x}\left(\frac{2xx}{x}+\frac{3}{x}\right)$$

Since $\frac{2xx}{x}$ and $\frac{3}{x}$ have the same denominator, add them by adding their numerators.

$$\frac{2x^{2}-3}{x}\times \frac{2xx+3}{x}$$

Do the multiplications in $2xx+3$.

$$\frac{2x^{2}-3}{x}\times \frac{2x^{2}+3}{x}$$

Multiply $\frac{2x^{2}-3}{x}$ times $\frac{2x^{2}+3}{x}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(2x^{2}-3\right)\left(2x^{2}+3\right)}{xx}$$

Multiply $x$ and $x$ to get $x^{2}$.

$$\frac{\left(2x^{2}-3\right)\left(2x^{2}+3\right)}{x^{2}}$$

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

$$\frac{\left(2x^{2}\right)^{2}-3^{2}}{x^{2}}$$

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

$$\frac{2^{2}\left(x^{2}\right)^{2}-3^{2}}{x^{2}}$$

To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.

$$\frac{2^{2}x^{4}-3^{2}}{x^{2}}$$

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

$$\frac{4x^{4}-3^{2}}{x^{2}}$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{4x^{4}-9}{x^{2}}$$

Expand

$4x^{2}-\frac{9}{x^{2}}$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply $2x$ times $\frac{x}{x}$.

$$\left(\frac{2xx}{x}-\frac{3}{x}\right)\left(2x+\frac{3}{x}\right)$$

Since $\frac{2xx}{x}$ and $\frac{3}{x}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{2xx-3}{x}\left(2x+\frac{3}{x}\right)$$

Do the multiplications in $2xx-3$.

$$\frac{2x^{2}-3}{x}\left(2x+\frac{3}{x}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply $2x$ times $\frac{x}{x}$.

$$\frac{2x^{2}-3}{x}\left(\frac{2xx}{x}+\frac{3}{x}\right)$$

Since $\frac{2xx}{x}$ and $\frac{3}{x}$ have the same denominator, add them by adding their numerators.

$$\frac{2x^{2}-3}{x}\times \frac{2xx+3}{x}$$

Do the multiplications in $2xx+3$.

$$\frac{2x^{2}-3}{x}\times \frac{2x^{2}+3}{x}$$

Multiply $\frac{2x^{2}-3}{x}$ times $\frac{2x^{2}+3}{x}$ by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(2x^{2}-3\right)\left(2x^{2}+3\right)}{xx}$$

Multiply $x$ and $x$ to get $x^{2}$.

$$\frac{\left(2x^{2}-3\right)\left(2x^{2}+3\right)}{x^{2}}$$

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

$$\frac{\left(2x^{2}\right)^{2}-3^{2}}{x^{2}}$$

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

$$\frac{2^{2}\left(x^{2}\right)^{2}-3^{2}}{x^{2}}$$

To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.

$$\frac{2^{2}x^{4}-3^{2}}{x^{2}}$$

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

$$\frac{4x^{4}-3^{2}}{x^{2}}$$

Calculate $3$ to the power of $2$ and get $9$.

$$\frac{4x^{4}-9}{x^{2}}$$