Solve x^2+4x=9(4/3)

$$x ^ { 2 } + 4 x = 9 ( 4 / 3 )$$

Solve for x

$$x=-6$$

$$x=2$$

Steps Using Factoring

Multiply 9 and \frac{4}{3} to get 12.

$$x^{2}+4x=12$$

Subtract 12 from both sides.

$$x^{2}+4x-12=0$$

To solve the equation, factor x^{2}+4x-12 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=4 ab=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-2\right)\left(x+6\right)$$

To find equation solutions, solve x-2=0 and x+6=0.

$$x=2 x=-6$$

Steps Using Factoring By Grouping

Multiply 9 and \frac{4}{3} to get 12.

$$x^{2}+4x=12$$

Subtract 12 from both sides.

$$x^{2}+4x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=-2 b=6$$

Rewrite x^{2}+4x-12 as \left(x^{2}-2x\right)+\left(6x-12\right).

$$\left(x^{2}-2x\right)+\left(6x-12\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-2\right)+6\left(x-2\right)$$

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

$$\left(x-2\right)\left(x+6\right)$$

To find equation solutions, solve x-2=0 and x+6=0.

$$x=2 x=-6$$

Steps Using the Quadratic Formula

Multiply 9 and \frac{4}{3} to get 12.

$$x^{2}+4x=12$$

Subtract 12 from both sides.

$$x^{2}+4x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

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

Multiply -4 times -12.

$$x=\frac{-4±\sqrt{16+48}}{2}$$

Add 16 to 48.

$$x=\frac{-4±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-4±8}{2}$$

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

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

Divide 4 by 2.

$$x=2$$

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

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

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=2 x=-6$$

Steps for Completing the Square

Multiply 9 and \frac{4}{3} to get 12.

$$x^{2}+4x=12$$

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

$$x^{2}+4x+2^{2}=12+2^{2}$$

Square 2.

$$x^{2}+4x+4=12+4$$

Add 12 to 4.

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

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{16}$$

Simplify.

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

Subtract 2 from both sides of the equation.

$$x=2 x=-6$$