Solve factorisation x(x - 1) + 3(x - 1) = 5 | 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

$$x ( x - 1 ) + 3 ( x - 1 ) = 5$$

Solve for x

$x=-4$
$x=2$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$x^{2}-x+3\left(x-1\right)=5$$

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

$$x^{2}-x+3x-3=5$$

Combine $-x$ and $3x$ to get $2x$.

$$x^{2}+2x-3=5$$

Subtract $5$ from both sides.

$$x^{2}+2x-3-5=0$$

Subtract $5$ from $-3$ to get $-8$.

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

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

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

Square $2$.

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

Multiply $-4$ times $-8$.

$$x=\frac{-2±\sqrt{4+32}}{2}$$

Add $4$ to $32$.

$$x=\frac{-2±\sqrt{36}}{2}$$

Take the square root of $36$.

$$x=\frac{-2±6}{2}$$

Now solve the equation $x=\frac{-2±6}{2}$ when $±$ is plus. Add $-2$ to $6$.

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

Divide $4$ by $2$.

$$x=2$$

Now solve the equation $x=\frac{-2±6}{2}$ when $±$ is minus. Subtract $6$ from $-2$.

$$x=-\frac{8}{2}$$

Divide $-8$ by $2$.

$$x=-4$$

The equation is now solved.

$$x=2$$ $$x=-4$$

Steps for Completing the Square

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

$$x^{2}-x+3\left(x-1\right)=5$$

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

$$x^{2}-x+3x-3=5$$

Combine $-x$ and $3x$ to get $2x$.

$$x^{2}+2x-3=5$$

Add $3$ to both sides.

$$x^{2}+2x=5+3$$

Add $5$ and $3$ to get $8$.

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

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.

$$x^{2}+2x+1^{2}=8+1^{2}$$

Square $1$.

$$x^{2}+2x+1=8+1$$

Add $8$ to $1$.

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

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

$$\left(x+1\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+1=3$$ $$x+1=-3$$

Subtract $1$ from both sides of the equation.

$$x=2$$ $$x=-4$$