Solve (1+4x)^2=81

$$( 1 + 4 x ) ^ { 2 } = 81$$

Solve for x

$$x=2$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Steps Using Factoring By Grouping

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

$$1+8x+16x^{2}=81$$

Subtract 81 from both sides.

$$1+8x+16x^{2}-81=0$$

Subtract 81 from 1 to get -80.

$$-80+8x+16x^{2}=0$$

Divide both sides by 8.

$$-10+x+2x^{2}=0$$

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

$$2x^{2}+x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=1 ab=2\left(-10\right)=-20$$

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 -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite 2x^{2}+x-10 as \left(2x^{2}-4x\right)+\left(5x-10\right).

$$\left(2x^{2}-4x\right)+\left(5x-10\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(x-2\right)+5\left(x-2\right)$$

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

$$\left(x-2\right)\left(2x+5\right)$$

To find equation solutions, solve x-2=0 and 2x+5=0.

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

Steps Using the Quadratic Formula

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

$$1+8x+16x^{2}=81$$

Subtract 81 from both sides.

$$1+8x+16x^{2}-81=0$$

Subtract 81 from 1 to get -80.

$$-80+8x+16x^{2}=0$$

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.

$$16x^{2}+8x-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 8 for b, and -80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 16\left(-80\right)}}{2\times 16}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 16\left(-80\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-8±\sqrt{64-64\left(-80\right)}}{2\times 16}$$

Multiply -64 times -80.

$$x=\frac{-8±\sqrt{64+5120}}{2\times 16}$$

Add 64 to 5120.

$$x=\frac{-8±\sqrt{5184}}{2\times 16}$$

Take the square root of 5184.

$$x=\frac{-8±72}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-8±72}{32}$$

Now solve the equation x=\frac{-8±72}{32} when ± is plus. Add -8 to 72.

$$x=\frac{64}{32}$$

Divide 64 by 32.

$$x=2$$

Now solve the equation x=\frac{-8±72}{32} when ± is minus. Subtract 72 from -8.

$$x=-\frac{80}{32}$$

Reduce the fraction \frac{-80}{32} to lowest terms by extracting and canceling out 16.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$1+8x+16x^{2}=81$$

Subtract 1 from both sides.

$$8x+16x^{2}=81-1$$

Subtract 1 from 81 to get 80.

$$8x+16x^{2}=80$$

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.

$$16x^{2}+8x=80$$

Divide both sides by 16.

$$\frac{16x^{2}+8x}{16}=\frac{80}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{8}{16}x=\frac{80}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{1}{2}x=\frac{80}{16}$$

Divide 80 by 16.

$$x^{2}+\frac{1}{2}x=5$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=5+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=5+\frac{1}{16}$$

Add 5 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{81}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{9}{4} x+\frac{1}{4}=-\frac{9}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

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