Solve and (A - 8)(B - c) | 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

$$\left. \begin{array} { l } { \frac { x - 6 } { x - 8 } 2 x = 3 } \\ { \{ n - 8 , 3 x + ( 8 ) = 3 , 3 } \end{array} \right.$$

Solve for x

$x = \frac{\sqrt{33} + 15}{4} \approx 5.186140662$
$x = \frac{15 - \sqrt{33}}{4} \approx 2.313859338$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $8$ since division by zero is not defined. Multiply both sides of the equation by $x-8$.

$$\left(x-6\right)\times 2x=3\left(x-8\right)$$

Use the distributive property to multiply $x-6$ by $2$.

$$\left(2x-12\right)x=3\left(x-8\right)$$

Use the distributive property to multiply $2x-12$ by $x$.

$$2x^{2}-12x=3\left(x-8\right)$$

Use the distributive property to multiply $3$ by $x-8$.

$$2x^{2}-12x=3x-24$$

Subtract $3x$ from both sides.

$$2x^{2}-12x-3x=-24$$

Combine $-12x$ and $-3x$ to get $-15x$.

$$2x^{2}-15x=-24$$

Add $24$ to both sides.

$$2x^{2}-15x+24=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-15$ for $b$, and $24$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square $-15$.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 2\times 24}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-15\right)±\sqrt{225-8\times 24}}{2\times 2}$$

Multiply $-8$ times $24$.

$$x=\frac{-\left(-15\right)±\sqrt{225-192}}{2\times 2}$$

Add $225$ to $-192$.

$$x=\frac{-\left(-15\right)±\sqrt{33}}{2\times 2}$$

The opposite of $-15$ is $15$.

$$x=\frac{15±\sqrt{33}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{15±\sqrt{33}}{4}$$

Now solve the equation $x=\frac{15±\sqrt{33}}{4}$ when $±$ is plus. Add $15$ to $\sqrt{33}$.

$$x=\frac{\sqrt{33}+15}{4}$$

Now solve the equation $x=\frac{15±\sqrt{33}}{4}$ when $±$ is minus. Subtract $\sqrt{33}$ from $15$.

$$x=\frac{15-\sqrt{33}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+15}{4}$$ $$x=\frac{15-\sqrt{33}}{4}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $8$ since division by zero is not defined. Multiply both sides of the equation by $x-8$.

$$\left(x-6\right)\times 2x=3\left(x-8\right)$$

Use the distributive property to multiply $x-6$ by $2$.

$$\left(2x-12\right)x=3\left(x-8\right)$$

Use the distributive property to multiply $2x-12$ by $x$.

$$2x^{2}-12x=3\left(x-8\right)$$

Use the distributive property to multiply $3$ by $x-8$.

$$2x^{2}-12x=3x-24$$

Subtract $3x$ from both sides.

$$2x^{2}-12x-3x=-24$$

Combine $-12x$ and $-3x$ to get $-15x$.

$$2x^{2}-15x=-24$$

Divide both sides by $2$.

$$\frac{2x^{2}-15x}{2}=-\frac{24}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}-\frac{15}{2}x=-\frac{24}{2}$$

Divide $-24$ by $2$.

$$x^{2}-\frac{15}{2}x=-12$$

Divide $-\frac{15}{2}$, the coefficient of the $x$ term, by $2$ to get $-\frac{15}{4}$. Then add the square of $-\frac{15}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{2}x+\left(-\frac{15}{4}\right)^{2}=-12+\left(-\frac{15}{4}\right)^{2}$$

Square $-\frac{15}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=-12+\frac{225}{16}$$

Add $-12$ to $\frac{225}{16}$.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{33}{16}$$

Factor $x^{2}-\frac{15}{2}x+\frac{225}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{15}{4}\right)^{2}=\frac{33}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{4}\right)^{2}}=\sqrt{\frac{33}{16}}$$

Simplify.

$$x-\frac{15}{4}=\frac{\sqrt{33}}{4}$$ $$x-\frac{15}{4}=-\frac{\sqrt{33}}{4}$$

Add $\frac{15}{4}$ to both sides of the equation.

$$x=\frac{\sqrt{33}+15}{4}$$ $$x=\frac{15-\sqrt{33}}{4}$$