Solve (4x+1)^2+9(4x+1)=-18

$$( 4 x + 1 ) ^ { 2 } + 9 ( 4 x + 1 ) = - 18$$

Solve for x

$$x = -\frac{7}{4} = -1\frac{3}{4} = -1.75$$

$$x=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1+9\left(4x+1\right)=-18$$

Use the distributive property to multiply 9 by 4x+1.

$$16x^{2}+8x+1+36x+9=-18$$

Combine 8x and 36x to get 44x.

$$16x^{2}+44x+1+9=-18$$

Add 1 and 9 to get 10.

$$16x^{2}+44x+10=-18$$

Add 18 to both sides.

$$16x^{2}+44x+10+18=0$$

Add 10 and 18 to get 28.

$$16x^{2}+44x+28=0$$

Divide both sides by 4.

$$4x^{2}+11x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=11 ab=4\times 7=28$$

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 28.

$$1,28 2,14 4,7$$

Calculate the sum for each pair.

$$1+28=29 2+14=16 4+7=11$$

The solution is the pair that gives sum 11.

$$a=4 b=7$$

Rewrite 4x^{2}+11x+7 as \left(4x^{2}+4x\right)+\left(7x+7\right).

$$\left(4x^{2}+4x\right)+\left(7x+7\right)$$

Factor out 4x in the first and 7 in the second group.

$$4x\left(x+1\right)+7\left(x+1\right)$$

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

$$\left(x+1\right)\left(4x+7\right)$$

To find equation solutions, solve x+1=0 and 4x+7=0.

$$x=-1 x=-\frac{7}{4}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1+9\left(4x+1\right)=-18$$

Use the distributive property to multiply 9 by 4x+1.

$$16x^{2}+8x+1+36x+9=-18$$

Combine 8x and 36x to get 44x.

$$16x^{2}+44x+1+9=-18$$

Add 1 and 9 to get 10.

$$16x^{2}+44x+10=-18$$

Add 18 to both sides.

$$16x^{2}+44x+10+18=0$$

Add 10 and 18 to get 28.

$$16x^{2}+44x+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 44 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-44±\sqrt{44^{2}-4\times 16\times 28}}{2\times 16}$$

Square 44.

$$x=\frac{-44±\sqrt{1936-4\times 16\times 28}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-44±\sqrt{1936-64\times 28}}{2\times 16}$$

Multiply -64 times 28.

$$x=\frac{-44±\sqrt{1936-1792}}{2\times 16}$$

Add 1936 to -1792.

$$x=\frac{-44±\sqrt{144}}{2\times 16}$$

Take the square root of 144.

$$x=\frac{-44±12}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-44±12}{32}$$

Now solve the equation x=\frac{-44±12}{32} when ± is plus. Add -44 to 12.

$$x=-\frac{32}{32}$$

Divide -32 by 32.

$$x=-1$$

Now solve the equation x=\frac{-44±12}{32} when ± is minus. Subtract 12 from -44.

$$x=-\frac{56}{32}$$

Reduce the fraction \frac{-56}{32} to lowest terms by extracting and canceling out 8.

$$x=-\frac{7}{4}$$

The equation is now solved.

$$x=-1 x=-\frac{7}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1+9\left(4x+1\right)=-18$$

Use the distributive property to multiply 9 by 4x+1.

$$16x^{2}+8x+1+36x+9=-18$$

Combine 8x and 36x to get 44x.

$$16x^{2}+44x+1+9=-18$$

Add 1 and 9 to get 10.

$$16x^{2}+44x+10=-18$$

Subtract 10 from both sides.

$$16x^{2}+44x=-18-10$$

Subtract 10 from -18 to get -28.

$$16x^{2}+44x=-28$$

Divide both sides by 16.

$$\frac{16x^{2}+44x}{16}=-\frac{28}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{44}{16}x=-\frac{28}{16}$$

Reduce the fraction \frac{44}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{11}{4}x=-\frac{28}{16}$$

Reduce the fraction \frac{-28}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{11}{4}x=-\frac{7}{4}$$

Divide \frac{11}{4}, the coefficient of the x term, by 2 to get \frac{11}{8}. Then add the square of \frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{4}x+\left(\frac{11}{8}\right)^{2}=-\frac{7}{4}+\left(\frac{11}{8}\right)^{2}$$

Square \frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{4}x+\frac{121}{64}=-\frac{7}{4}+\frac{121}{64}$$

Add -\frac{7}{4} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{4}x+\frac{121}{64}=\frac{9}{64}$$

Factor x^{2}+\frac{11}{4}x+\frac{121}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{8}\right)^{2}=\frac{9}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{8}\right)^{2}}=\sqrt{\frac{9}{64}}$$

Simplify.

$$x+\frac{11}{8}=\frac{3}{8} x+\frac{11}{8}=-\frac{3}{8}$$

Subtract \frac{11}{8} from both sides of the equation.

$$x=-1 x=-\frac{7}{4}$$