How to solve 4(2x+1)=(x)^(2)

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.

Solve 4(2x+1)=(x)^(2) | Equatix

Question

$$4(2x+1)= { x }^{ 2 }$$

Solve for x

$x=2\sqrt{5}+4\approx 8.472135955$
$x=4-2\sqrt{5}\approx -0.472135955$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $4$ by $2x+1$.

$$8x+4=x^{2}$$

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

$$8x+4-x^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-x^{2}+8x+4=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $8$ for $b$, and $4$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square $8$.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-8±\sqrt{64+4\times 4}}{2\left(-1\right)}$$

Multiply $4$ times $4$.

$$x=\frac{-8±\sqrt{64+16}}{2\left(-1\right)}$$

Add $64$ to $16$.

$$x=\frac{-8±\sqrt{80}}{2\left(-1\right)}$$

Take the square root of $80$.

$$x=\frac{-8±4\sqrt{5}}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{-8±4\sqrt{5}}{-2}$$

Now solve the equation $x=\frac{-8±4\sqrt{5}}{-2}$ when $±$ is plus. Add $-8$ to $4\sqrt{5}$.

$$x=\frac{4\sqrt{5}-8}{-2}$$

Divide $-8+4\sqrt{5}$ by $-2$.

$$x=4-2\sqrt{5}$$

Now solve the equation $x=\frac{-8±4\sqrt{5}}{-2}$ when $±$ is minus. Subtract $4\sqrt{5}$ from $-8$.

$$x=\frac{-4\sqrt{5}-8}{-2}$$

Divide $-8-4\sqrt{5}$ by $-2$.

$$x=2\sqrt{5}+4$$

The equation is now solved.

$$x=4-2\sqrt{5}$$ $$x=2\sqrt{5}+4$$

Steps for Completing the Square

Use the distributive property to multiply $4$ by $2x+1$.

$$8x+4=x^{2}$$

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

$$8x+4-x^{2}=0$$

Subtract $4$ from both sides. Anything subtracted from zero gives its negation.

$$8x-x^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$-x^{2}+8x=-4$$

Divide both sides by $-1$.

$$\frac{-x^{2}+8x}{-1}=-\frac{4}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\frac{8}{-1}x=-\frac{4}{-1}$$

Divide $8$ by $-1$.

$$x^{2}-8x=-\frac{4}{-1}$$

Divide $-4$ by $-1$.

$$x^{2}-8x=4$$

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

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

Square $-4$.

$$x^{2}-8x+16=4+16$$

Add $4$ to $16$.

$$x^{2}-8x+16=20$$

Factor $x^{2}-8x+16$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-4\right)^{2}=20$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{20}$$

Simplify.

$$x-4=2\sqrt{5}$$ $$x-4=-2\sqrt{5}$$

Add $4$ to both sides of the equation.

$$x=2\sqrt{5}+4$$ $$x=4-2\sqrt{5}$$

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Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square