Solve 128(1+x)(1+x)=200

$$128(1+x)(1+x)=200$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x = -\frac{9}{4} = -2\frac{1}{4} = -2.25$$

Steps Using the Quadratic Formula

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$128\left(1+x\right)^{2}=200$$

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

$$128\left(1+2x+x^{2}\right)=200$$

Use the distributive property to multiply 128 by 1+2x+x^{2}.

$$128+256x+128x^{2}=200$$

Subtract 200 from both sides.

$$128+256x+128x^{2}-200=0$$

Subtract 200 from 128 to get -72.

$$-72+256x+128x^{2}=0$$

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.

$$128x^{2}+256x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 128 for a, 256 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-256±\sqrt{256^{2}-4\times 128\left(-72\right)}}{2\times 128}$$

Square 256.

$$x=\frac{-256±\sqrt{65536-4\times 128\left(-72\right)}}{2\times 128}$$

Multiply -4 times 128.

$$x=\frac{-256±\sqrt{65536-512\left(-72\right)}}{2\times 128}$$

Multiply -512 times -72.

$$x=\frac{-256±\sqrt{65536+36864}}{2\times 128}$$

Add 65536 to 36864.

$$x=\frac{-256±\sqrt{102400}}{2\times 128}$$

Take the square root of 102400.

$$x=\frac{-256±320}{2\times 128}$$

Multiply 2 times 128.

$$x=\frac{-256±320}{256}$$

Now solve the equation x=\frac{-256±320}{256} when ± is plus. Add -256 to 320.

$$x=\frac{64}{256}$$

Reduce the fraction \frac{64}{256} to lowest terms by extracting and canceling out 64.

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

Now solve the equation x=\frac{-256±320}{256} when ± is minus. Subtract 320 from -256.

$$x=-\frac{576}{256}$$

Reduce the fraction \frac{-576}{256} to lowest terms by extracting and canceling out 64.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$128\left(1+x\right)^{2}=200$$

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

$$128\left(1+2x+x^{2}\right)=200$$

Use the distributive property to multiply 128 by 1+2x+x^{2}.

$$128+256x+128x^{2}=200$$

Subtract 128 from both sides.

$$256x+128x^{2}=200-128$$

Subtract 128 from 200 to get 72.

$$256x+128x^{2}=72$$

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.

$$128x^{2}+256x=72$$

Divide both sides by 128.

$$\frac{128x^{2}+256x}{128}=\frac{72}{128}$$

Dividing by 128 undoes the multiplication by 128.

$$x^{2}+\frac{256}{128}x=\frac{72}{128}$$

Divide 256 by 128.

$$x^{2}+2x=\frac{72}{128}$$

Reduce the fraction \frac{72}{128} to lowest terms by extracting and canceling out 8.

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

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

$$x^{2}+2x+1^{2}=\frac{9}{16}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{9}{16}+1$$

Add \frac{9}{16} to 1.

$$x^{2}+2x+1=\frac{25}{16}$$

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

$$\left(x+1\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+1=\frac{5}{4} x+1=-\frac{5}{4}$$

Subtract 1 from both sides of the equation.

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