Solve 2+(3)/((2x+1)(2x-1))=3 | 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

$$2+ \frac{ 3 }{ (2x+1)(2x-1) } =3$$

Solve for x

$x=-1$
$x=1$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

Variable $x$ cannot be equal to any of the values $-\frac{1}{2},\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $\left(2x-1\right)\left(2x+1\right)$.

$$\left(2x-1\right)\left(2x+1\right)\times 2+3=3\left(2x-1\right)\left(2x+1\right)$$

Consider $\left(2x-1\right)\left(2x+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

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

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

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

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

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

Use the distributive property to multiply $4x^{2}-1$ by $2$.

$$8x^{2}-2+3=3\left(2x-1\right)\left(2x+1\right)$$

Add $-2$ and $3$ to get $1$.

$$8x^{2}+1=3\left(2x-1\right)\left(2x+1\right)$$

Use the distributive property to multiply $3$ by $2x-1$.

$$8x^{2}+1=\left(6x-3\right)\left(2x+1\right)$$

Use the distributive property to multiply $6x-3$ by $2x+1$ and combine like terms.

$$8x^{2}+1=12x^{2}-3$$

Subtract $12x^{2}$ from both sides.

$$8x^{2}+1-12x^{2}=-3$$

Combine $8x^{2}$ and $-12x^{2}$ to get $-4x^{2}$.

$$-4x^{2}+1=-3$$

Subtract $1$ from both sides.

$$-4x^{2}=-3-1$$

Subtract $1$ from $-3$ to get $-4$.

$$-4x^{2}=-4$$

Divide both sides by $-4$.

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

Divide $-4$ by $-4$ to get $1$.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1$$ $$x=-1$$

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-\frac{1}{2},\frac{1}{2}$ since division by zero is not defined. Multiply both sides of the equation by $\left(2x-1\right)\left(2x+1\right)$.

$$\left(2x-1\right)\left(2x+1\right)\times 2+3=3\left(2x-1\right)\left(2x+1\right)$$

Consider $\left(2x-1\right)\left(2x+1\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.

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

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

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

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

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

Use the distributive property to multiply $4x^{2}-1$ by $2$.

$$8x^{2}-2+3=3\left(2x-1\right)\left(2x+1\right)$$

Add $-2$ and $3$ to get $1$.

$$8x^{2}+1=3\left(2x-1\right)\left(2x+1\right)$$

Use the distributive property to multiply $3$ by $2x-1$.

$$8x^{2}+1=\left(6x-3\right)\left(2x+1\right)$$

Use the distributive property to multiply $6x-3$ by $2x+1$ and combine like terms.

$$8x^{2}+1=12x^{2}-3$$

Subtract $12x^{2}$ from both sides.

$$8x^{2}+1-12x^{2}=-3$$

Combine $8x^{2}$ and $-12x^{2}$ to get $-4x^{2}$.

$$-4x^{2}+1=-3$$

Add $3$ to both sides.

$$-4x^{2}+1+3=0$$

Add $1$ and $3$ to get $4$.

$$-4x^{2}+4=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-4$ for $a$, $0$ for $b$, and $4$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)\times 4}}{2\left(-4\right)}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\left(-4\right)\times 4}}{2\left(-4\right)}$$

Multiply $-4$ times $-4$.

$$x=\frac{0±\sqrt{16\times 4}}{2\left(-4\right)}$$

Multiply $16$ times $4$.

$$x=\frac{0±\sqrt{64}}{2\left(-4\right)}$$

Take the square root of $64$.

$$x=\frac{0±8}{2\left(-4\right)}$$

Multiply $2$ times $-4$.

$$x=\frac{0±8}{-8}$$

Now solve the equation $x=\frac{0±8}{-8}$ when $±$ is plus. Divide $8$ by $-8$.

$$x=-1$$

Now solve the equation $x=\frac{0±8}{-8}$ when $±$ is minus. Divide $-8$ by $-8$.

$$x=1$$

The equation is now solved.

$$x=-1$$ $$x=1$$