Solve 0=4x^2+12x+9/2x^2-x-4

$$0= \frac{ 4 { x }^{ 2 } +12x+9 }{ 2 { x }^{ 2 } -x } -4$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values 0,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by x\left(2x-1\right).

$$0=4x^{2}+12x+9+x\left(2x-1\right)\left(-4\right)$$

Use the distributive property to multiply x by 2x-1.

$$0=4x^{2}+12x+9+\left(2x^{2}-x\right)\left(-4\right)$$

Use the distributive property to multiply 2x^{2}-x by -4.

$$0=4x^{2}+12x+9-8x^{2}+4x$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$0=-4x^{2}+12x+9+4x$$

Combine 12x and 4x to get 16x.

$$0=-4x^{2}+16x+9$$

Swap sides so that all variable terms are on the left hand side.

$$-4x^{2}+16x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=16 ab=-4\times 9=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 16.

$$a=18 b=-2$$

Rewrite -4x^{2}+16x+9 as \left(-4x^{2}+18x\right)+\left(-2x+9\right).

$$\left(-4x^{2}+18x\right)+\left(-2x+9\right)$$

Factor out -2x in the first and -1 in the second group.

$$-2x\left(2x-9\right)-\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(-2x-1\right)$$

To find equation solutions, solve 2x-9=0 and -2x-1=0.

$$x=\frac{9}{2} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by x\left(2x-1\right).

$$0=4x^{2}+12x+9+x\left(2x-1\right)\left(-4\right)$$

Use the distributive property to multiply x by 2x-1.

$$0=4x^{2}+12x+9+\left(2x^{2}-x\right)\left(-4\right)$$

Use the distributive property to multiply 2x^{2}-x by -4.

$$0=4x^{2}+12x+9-8x^{2}+4x$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$0=-4x^{2}+12x+9+4x$$

Combine 12x and 4x to get 16x.

$$0=-4x^{2}+16x+9$$

Swap sides so that all variable terms are on the left hand side.

$$-4x^{2}+16x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 16 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-4\right)\times 9}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-16±\sqrt{256+16\times 9}}{2\left(-4\right)}$$

Multiply 16 times 9.

$$x=\frac{-16±\sqrt{256+144}}{2\left(-4\right)}$$

Add 256 to 144.

$$x=\frac{-16±\sqrt{400}}{2\left(-4\right)}$$

Take the square root of 400.

$$x=\frac{-16±20}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-16±20}{-8}$$

Now solve the equation x=\frac{-16±20}{-8} when ± is plus. Add -16 to 20.

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

Reduce the fraction \frac{4}{-8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-16±20}{-8} when ± is minus. Subtract 20 from -16.

$$x=-\frac{36}{-8}$$

Reduce the fraction \frac{-36}{-8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

$$x=-\frac{1}{2} x=\frac{9}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by x\left(2x-1\right).

$$0=4x^{2}+12x+9+x\left(2x-1\right)\left(-4\right)$$

Use the distributive property to multiply x by 2x-1.

$$0=4x^{2}+12x+9+\left(2x^{2}-x\right)\left(-4\right)$$

Use the distributive property to multiply 2x^{2}-x by -4.

$$0=4x^{2}+12x+9-8x^{2}+4x$$

Combine 4x^{2} and -8x^{2} to get -4x^{2}.

$$0=-4x^{2}+12x+9+4x$$

Combine 12x and 4x to get 16x.

$$0=-4x^{2}+16x+9$$

Swap sides so that all variable terms are on the left hand side.

$$-4x^{2}+16x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-4x^{2}+16x=-9$$

Divide both sides by -4.

$$\frac{-4x^{2}+16x}{-4}=-\frac{9}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{16}{-4}x=-\frac{9}{-4}$$

Divide 16 by -4.

$$x^{2}-4x=-\frac{9}{-4}$$

Divide -9 by -4.

$$x^{2}-4x=\frac{9}{4}$$

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.

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

Square -2.

$$x^{2}-4x+4=\frac{9}{4}+4$$

Add \frac{9}{4} to 4.

$$x^{2}-4x+4=\frac{25}{4}$$

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}.

$$\left(x-2\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-2=\frac{5}{2} x-2=-\frac{5}{2}$$

Add 2 to both sides of the equation.

$$x=\frac{9}{2} x=-\frac{1}{2}$$