Solve 4x^2+24x+16=5

$$4 x ^ { 2 } + 24 x + 16 = 5$$

Solve for x

$$x = -\frac{11}{2} = -5\frac{1}{2} = -5.5$$

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

Steps Using Factoring By Grouping

Subtract 5 from both sides.

$$4x^{2}+24x+16-5=0$$

Subtract 5 from 16 to get 11.

$$4x^{2}+24x+11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+11. To find a and b, set up a system to be solved.

$$a+b=24 ab=4\times 11=44$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 44.

$$1,44 2,22 4,11$$

Calculate the sum for each pair.

$$1+44=45 2+22=24 4+11=15$$

The solution is the pair that gives sum 24.

$$a=2 b=22$$

Rewrite 4x^{2}+24x+11 as \left(4x^{2}+2x\right)+\left(22x+11\right).

$$\left(4x^{2}+2x\right)+\left(22x+11\right)$$

Factor out 2x in the first and 11 in the second group.

$$2x\left(2x+1\right)+11\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(2x+11\right)$$

To find equation solutions, solve 2x+1=0 and 2x+11=0.

$$x=-\frac{1}{2} x=-\frac{11}{2}$$

Steps Using the Quadratic Formula

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.

$$4x^{2}+24x+16=5$$

Subtract 5 from both sides of the equation.

$$4x^{2}+24x+16-5=5-5$$

Subtracting 5 from itself leaves 0.

$$4x^{2}+24x+16-5=0$$

Subtract 5 from 16.

$$4x^{2}+24x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 24 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\times 4\times 11}}{2\times 4}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 4\times 11}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-24±\sqrt{576-16\times 11}}{2\times 4}$$

Multiply -16 times 11.

$$x=\frac{-24±\sqrt{576-176}}{2\times 4}$$

Add 576 to -176.

$$x=\frac{-24±\sqrt{400}}{2\times 4}$$

Take the square root of 400.

$$x=\frac{-24±20}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-24±20}{8}$$

Now solve the equation x=\frac{-24±20}{8} when ± is plus. Add -24 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{-24±20}{8} when ± is minus. Subtract 20 from -24.

$$x=-\frac{44}{8}$$

Reduce the fraction \frac{-44}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{11}{2}$$

The equation is now solved.

$$x=-\frac{1}{2} x=-\frac{11}{2}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$4x^{2}+24x+16=5$$

Subtract 16 from both sides of the equation.

$$4x^{2}+24x+16-16=5-16$$

Subtracting 16 from itself leaves 0.

$$4x^{2}+24x=5-16$$

Subtract 16 from 5.

$$4x^{2}+24x=-11$$

Divide both sides by 4.

$$\frac{4x^{2}+24x}{4}=-\frac{11}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{24}{4}x=-\frac{11}{4}$$

Divide 24 by 4.

$$x^{2}+6x=-\frac{11}{4}$$

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.

$$x^{2}+6x+3^{2}=-\frac{11}{4}+3^{2}$$

Square 3.

$$x^{2}+6x+9=-\frac{11}{4}+9$$

Add -\frac{11}{4} to 9.

$$x^{2}+6x+9=\frac{25}{4}$$

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}.

$$\left(x+3\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+3=\frac{5}{2} x+3=-\frac{5}{2}$$

Subtract 3 from both sides of the equation.

$$x=-\frac{1}{2} x=-\frac{11}{2}$$