Solve x^2-16/3x^2+12x=1

$$\frac { x ^ { 2 } - 16 } { 3 x ^ { 2 } + 12 x } = 1$$

Solve for x

$$x=-2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+4\right).

$$x^{2}-16=3x\left(x+4\right)$$

Use the distributive property to multiply 3x by x+4.

$$x^{2}-16=3x^{2}+12x$$

Subtract 3x^{2} from both sides.

$$x^{2}-16-3x^{2}=12x$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-16=12x$$

Subtract 12x from both sides.

$$-2x^{2}-16-12x=0$$

Divide both sides by 2.

$$-x^{2}-8-6x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-6x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=-\left(-8\right)=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-2 b=-4$$

Rewrite -x^{2}-6x-8 as \left(-x^{2}-2x\right)+\left(-4x-8\right).

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

Factor out x in the first and 4 in the second group.

$$x\left(-x-2\right)+4\left(-x-2\right)$$

Factor out common term -x-2 by using distributive property.

$$\left(-x-2\right)\left(x+4\right)$$

To find equation solutions, solve -x-2=0 and x+4=0.

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

Variable x cannot be equal to -4.

$$x=-2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+4\right).

$$x^{2}-16=3x\left(x+4\right)$$

Use the distributive property to multiply 3x by x+4.

$$x^{2}-16=3x^{2}+12x$$

Subtract 3x^{2} from both sides.

$$x^{2}-16-3x^{2}=12x$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-16=12x$$

Subtract 12x from both sides.

$$-2x^{2}-16-12x=0$$

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.

$$-2x^{2}-12x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -12 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-2\right)\left(-16\right)}}{2\left(-2\right)}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\left(-2\right)\left(-16\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-12\right)±\sqrt{144+8\left(-16\right)}}{2\left(-2\right)}$$

Multiply 8 times -16.

$$x=\frac{-\left(-12\right)±\sqrt{144-128}}{2\left(-2\right)}$$

Add 144 to -128.

$$x=\frac{-\left(-12\right)±\sqrt{16}}{2\left(-2\right)}$$

Take the square root of 16.

$$x=\frac{-\left(-12\right)±4}{2\left(-2\right)}$$

The opposite of -12 is 12.

$$x=\frac{12±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{12±4}{-4}$$

Now solve the equation x=\frac{12±4}{-4} when ± is plus. Add 12 to 4.

$$x=\frac{16}{-4}$$

Divide 16 by -4.

$$x=-4$$

Now solve the equation x=\frac{12±4}{-4} when ± is minus. Subtract 4 from 12.

$$x=\frac{8}{-4}$$

Divide 8 by -4.

$$x=-2$$

The equation is now solved.

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

Variable x cannot be equal to -4.

$$x=-2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x+4\right).

$$x^{2}-16=3x\left(x+4\right)$$

Use the distributive property to multiply 3x by x+4.

$$x^{2}-16=3x^{2}+12x$$

Subtract 3x^{2} from both sides.

$$x^{2}-16-3x^{2}=12x$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-16=12x$$

Subtract 12x from both sides.

$$-2x^{2}-16-12x=0$$

Add 16 to both sides. Anything plus zero gives itself.

$$-2x^{2}-12x=16$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{12}{-2}\right)x=\frac{16}{-2}$$

Divide -12 by -2.

$$x^{2}+6x=\frac{16}{-2}$$

Divide 16 by -2.

$$x^{2}+6x=-8$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-8+3^{2}$$

Square 3.

$$x^{2}+6x+9=-8+9$$

Add -8 to 9.

$$x^{2}+6x+9=1$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 3 from both sides of the equation.

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

Variable x cannot be equal to -4.

$$x=-2$$