Solve 4bx = 2 + sqrt(3) find the value cof * (x - 1/x) ^ 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

$$4 b x = 2 + \sqrt { 3 } ( x - \frac { 1 } { x } ) ^ { 2 }$$

Solve for b

$b=\frac{\sqrt{3}x^{4}-2\sqrt{3}x^{2}+2x^{2}+\sqrt{3}}{4x^{3}}$
$x\neq 0$

Steps for Solving Linear Equation

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

$$4bx=2+\sqrt{3}\left(\frac{xx}{x}-\frac{1}{x}\right)^{2}$$

Since $\frac{xx}{x}$ and $\frac{1}{x}$ have the same denominator, subtract them by subtracting their numerators.

$$4bx=2+\sqrt{3}\times \left(\frac{xx-1}{x}\right)^{2}$$

Do the multiplications in $xx-1$.

$$4bx=2+\sqrt{3}\times \left(\frac{x^{2}-1}{x}\right)^{2}$$

To raise $\frac{x^{2}-1}{x}$ to a power, raise both numerator and denominator to the power and then divide.

$$4bx=2+\sqrt{3}\times \frac{\left(x^{2}-1\right)^{2}}{x^{2}}$$

Express $\sqrt{3}\times \frac{\left(x^{2}-1\right)^{2}}{x^{2}}$ as a single fraction.

$$4bx=2+\frac{\sqrt{3}\left(x^{2}-1\right)^{2}}{x^{2}}$$

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

$$4bx=\frac{2x^{2}}{x^{2}}+\frac{\sqrt{3}\left(x^{2}-1\right)^{2}}{x^{2}}$$

Since $\frac{2x^{2}}{x^{2}}$ and $\frac{\sqrt{3}\left(x^{2}-1\right)^{2}}{x^{2}}$ have the same denominator, add them by adding their numerators.

$$4bx=\frac{2x^{2}+\sqrt{3}\left(x^{2}-1\right)^{2}}{x^{2}}$$

Do the multiplications in $2x^{2}+\sqrt{3}\left(x^{2}-1\right)^{2}$.

$$4bx=\frac{2x^{2}+\sqrt{3}x^{4}-2\sqrt{3}x^{2}+\sqrt{3}}{x^{2}}$$

Multiply both sides of the equation by $x^{2}$.

$$4bxx^{2}=2x^{2}+\sqrt{3}x^{4}-2\sqrt{3}x^{2}+\sqrt{3}$$

To multiply powers of the same base, add their exponents. Add $1$ and $2$ to get $3$.

$$4bx^{3}=2x^{2}+\sqrt{3}x^{4}-2\sqrt{3}x^{2}+\sqrt{3}$$

The equation is in standard form.

$$4x^{3}b=\sqrt{3}x^{4}-2\sqrt{3}x^{2}+2x^{2}+\sqrt{3}$$

Divide both sides by $4x^{3}$.

$$\frac{4x^{3}b}{4x^{3}}=\frac{\sqrt{3}x^{4}-2\sqrt{3}x^{2}+2x^{2}+\sqrt{3}}{4x^{3}}$$

Dividing by $4x^{3}$ undoes the multiplication by $4x^{3}$.

$$b=\frac{\sqrt{3}x^{4}-2\sqrt{3}x^{2}+2x^{2}+\sqrt{3}}{4x^{3}}$$