How to solve (x+3)(x-1)=3/2

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

Question

$$(x+3)(x-1)=3/2$$

Solve for x

$x=\frac{\sqrt{22}}{2}-1\approx 1.34520788$
$x=-\frac{\sqrt{22}}{2}-1\approx -3.34520788$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

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

Subtract $\frac{3}{2}$ from both sides.

$$x^{2}+2x-3-\frac{3}{2}=0$$

Subtract $\frac{3}{2}$ from $-3$ to get $-\frac{9}{2}$.

$$x^{2}+2x-\frac{9}{2}=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $2$ for $b$, and $-\frac{9}{2}$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-\frac{9}{2}\right)}}{2}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\left(-\frac{9}{2}\right)}}{2}$$

Multiply $-4$ times $-\frac{9}{2}$.

$$x=\frac{-2±\sqrt{4+18}}{2}$$

Add $4$ to $18$.

$$x=\frac{-2±\sqrt{22}}{2}$$

Now solve the equation $x=\frac{-2±\sqrt{22}}{2}$ when $±$ is plus. Add $-2$ to $\sqrt{22}$.

$$x=\frac{\sqrt{22}-2}{2}$$

Divide $-2+\sqrt{22}$ by $2$.

$$x=\frac{\sqrt{22}}{2}-1$$

Now solve the equation $x=\frac{-2±\sqrt{22}}{2}$ when $±$ is minus. Subtract $\sqrt{22}$ from $-2$.

$$x=\frac{-\sqrt{22}-2}{2}$$

Divide $-2-\sqrt{22}$ by $2$.

$$x=-\frac{\sqrt{22}}{2}-1$$

The equation is now solved.

$$x=\frac{\sqrt{22}}{2}-1$$ $$x=-\frac{\sqrt{22}}{2}-1$$

Steps for Completing the Square

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

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

Add $3$ to both sides.

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

Add $\frac{3}{2}$ and $3$ to get $\frac{9}{2}$.

$$x^{2}+2x=\frac{9}{2}$$

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

$$x^{2}+2x+1^{2}=\frac{9}{2}+1^{2}$$

Square $1$.

$$x^{2}+2x+1=\frac{9}{2}+1$$

Add $\frac{9}{2}$ to $1$.

$$x^{2}+2x+1=\frac{11}{2}$$

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

$$\left(x+1\right)^{2}=\frac{11}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{11}{2}}$$

Simplify.

$$x+1=\frac{\sqrt{22}}{2}$$ $$x+1=-\frac{\sqrt{22}}{2}$$

Subtract $1$ from both sides of the equation.

$$x=\frac{\sqrt{22}}{2}-1$$ $$x=-\frac{\sqrt{22}}{2}-1$$

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Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square