Solve h ) (2x + 1) ^ 2 = 3x - 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

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

Solve for x (complex solution)

$x=\frac{-1+\sqrt{47}i}{8}\approx -0.125+0.856956825i$
$x=\frac{-\sqrt{47}i-1}{8}\approx -0.125-0.856956825i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+1\right)^{2}$.

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

Subtract $3x$ from both sides.

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

Combine $4x$ and $-3x$ to get $x$.

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

Add $2$ to both sides.

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

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

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

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $4$ for $a$, $1$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\times 3}}{2\times 4}$$

Square $1$.

$$x=\frac{-1±\sqrt{1-4\times 4\times 3}}{2\times 4}$$

Multiply $-4$ times $4$.

$$x=\frac{-1±\sqrt{1-16\times 3}}{2\times 4}$$

Multiply $-16$ times $3$.

$$x=\frac{-1±\sqrt{1-48}}{2\times 4}$$

Add $1$ to $-48$.

$$x=\frac{-1±\sqrt{-47}}{2\times 4}$$

Take the square root of $-47$.

$$x=\frac{-1±\sqrt{47}i}{2\times 4}$$

Multiply $2$ times $4$.

$$x=\frac{-1±\sqrt{47}i}{8}$$

Now solve the equation $x=\frac{-1±\sqrt{47}i}{8}$ when $±$ is plus. Add $-1$ to $i\sqrt{47}$.

$$x=\frac{-1+\sqrt{47}i}{8}$$

Now solve the equation $x=\frac{-1±\sqrt{47}i}{8}$ when $±$ is minus. Subtract $i\sqrt{47}$ from $-1$.

$$x=\frac{-\sqrt{47}i-1}{8}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{47}i}{8}$$ $$x=\frac{-\sqrt{47}i-1}{8}$$

Steps for Completing the Square

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(2x+1\right)^{2}$.

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

Subtract $3x$ from both sides.

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

Combine $4x$ and $-3x$ to get $x$.

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

Subtract $1$ from both sides.

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

Subtract $1$ from $-2$ to get $-3$.

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

Divide both sides by $4$.

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

Dividing by $4$ undoes the multiplication by $4$.

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

Divide $\frac{1}{4}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{8}$. Then add the square of $\frac{1}{8}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=-\frac{3}{4}+\left(\frac{1}{8}\right)^{2}$$

Square $\frac{1}{8}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{3}{4}+\frac{1}{64}$$

Add $-\frac{3}{4}$ to $\frac{1}{64}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=-\frac{47}{64}$$

Factor $x^{2}+\frac{1}{4}x+\frac{1}{64}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{8}\right)^{2}=-\frac{47}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{-\frac{47}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{\sqrt{47}i}{8}$$ $$x+\frac{1}{8}=-\frac{\sqrt{47}i}{8}$$

Subtract $\frac{1}{8}$ from both sides of the equation.

$$x=\frac{-1+\sqrt{47}i}{8}$$ $$x=\frac{-\sqrt{47}i-1}{8}$$