Solve 9(x-2)^2-16(x+1)^2=0

$$9 ( x - 2 ) ^ { 2 } - 16 ( x + 1 ) ^ { 2 } = 0$$

Solve for x

$$x=\frac{2}{7}\approx 0.285714286$$

$$x=-10$$

Steps Using Factoring By Grouping

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

$$9\left(x^{2}-4x+4\right)-16\left(x+1\right)^{2}=0$$

Use the distributive property to multiply 9 by x^{2}-4x+4.

$$9x^{2}-36x+36-16\left(x+1\right)^{2}=0$$

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

$$9x^{2}-36x+36-16\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -16 by x^{2}+2x+1.

$$9x^{2}-36x+36-16x^{2}-32x-16=0$$

Combine 9x^{2} and -16x^{2} to get -7x^{2}.

$$-7x^{2}-36x+36-32x-16=0$$

Combine -36x and -32x to get -68x.

$$-7x^{2}-68x+36-16=0$$

Subtract 16 from 36 to get 20.

$$-7x^{2}-68x+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -7x^{2}+ax+bx+20. To find a and b, set up a system to be solved.

$$a+b=-68 ab=-7\times 20=-140$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -140.

$$1,-140 2,-70 4,-35 5,-28 7,-20 10,-14$$

Calculate the sum for each pair.

$$1-140=-139 2-70=-68 4-35=-31 5-28=-23 7-20=-13 10-14=-4$$

The solution is the pair that gives sum -68.

$$a=2 b=-70$$

Rewrite -7x^{2}-68x+20 as \left(-7x^{2}+2x\right)+\left(-70x+20\right).

$$\left(-7x^{2}+2x\right)+\left(-70x+20\right)$$

Factor out -x in the first and -10 in the second group.

$$-x\left(7x-2\right)-10\left(7x-2\right)$$

Factor out common term 7x-2 by using distributive property.

$$\left(7x-2\right)\left(-x-10\right)$$

To find equation solutions, solve 7x-2=0 and -x-10=0.

$$x=\frac{2}{7} x=-10$$

Steps Using the Quadratic Formula

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

$$9\left(x^{2}-4x+4\right)-16\left(x+1\right)^{2}=0$$

Use the distributive property to multiply 9 by x^{2}-4x+4.

$$9x^{2}-36x+36-16\left(x+1\right)^{2}=0$$

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

$$9x^{2}-36x+36-16\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -16 by x^{2}+2x+1.

$$9x^{2}-36x+36-16x^{2}-32x-16=0$$

Combine 9x^{2} and -16x^{2} to get -7x^{2}.

$$-7x^{2}-36x+36-32x-16=0$$

Combine -36x and -32x to get -68x.

$$-7x^{2}-68x+36-16=0$$

Subtract 16 from 36 to get 20.

$$-7x^{2}-68x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, -68 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-68\right)±\sqrt{\left(-68\right)^{2}-4\left(-7\right)\times 20}}{2\left(-7\right)}$$

Square -68.

$$x=\frac{-\left(-68\right)±\sqrt{4624-4\left(-7\right)\times 20}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-\left(-68\right)±\sqrt{4624+28\times 20}}{2\left(-7\right)}$$

Multiply 28 times 20.

$$x=\frac{-\left(-68\right)±\sqrt{4624+560}}{2\left(-7\right)}$$

Add 4624 to 560.

$$x=\frac{-\left(-68\right)±\sqrt{5184}}{2\left(-7\right)}$$

Take the square root of 5184.

$$x=\frac{-\left(-68\right)±72}{2\left(-7\right)}$$

The opposite of -68 is 68.

$$x=\frac{68±72}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{68±72}{-14}$$

Now solve the equation x=\frac{68±72}{-14} when ± is plus. Add 68 to 72.

$$x=\frac{140}{-14}$$

Divide 140 by -14.

$$x=-10$$

Now solve the equation x=\frac{68±72}{-14} when ± is minus. Subtract 72 from 68.

$$x=-\frac{4}{-14}$$

Reduce the fraction \frac{-4}{-14} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{7}$$

The equation is now solved.

$$x=-10 x=\frac{2}{7}$$

Steps for Completing the Square

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

$$9\left(x^{2}-4x+4\right)-16\left(x+1\right)^{2}=0$$

Use the distributive property to multiply 9 by x^{2}-4x+4.

$$9x^{2}-36x+36-16\left(x+1\right)^{2}=0$$

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

$$9x^{2}-36x+36-16\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -16 by x^{2}+2x+1.

$$9x^{2}-36x+36-16x^{2}-32x-16=0$$

Combine 9x^{2} and -16x^{2} to get -7x^{2}.

$$-7x^{2}-36x+36-32x-16=0$$

Combine -36x and -32x to get -68x.

$$-7x^{2}-68x+36-16=0$$

Subtract 16 from 36 to get 20.

$$-7x^{2}-68x+20=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-7x^{2}-68x=-20$$

Divide both sides by -7.

$$\frac{-7x^{2}-68x}{-7}=-\frac{20}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\left(-\frac{68}{-7}\right)x=-\frac{20}{-7}$$

Divide -68 by -7.

$$x^{2}+\frac{68}{7}x=-\frac{20}{-7}$$

Divide -20 by -7.

$$x^{2}+\frac{68}{7}x=\frac{20}{7}$$

Divide \frac{68}{7}, the coefficient of the x term, by 2 to get \frac{34}{7}. Then add the square of \frac{34}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{68}{7}x+\left(\frac{34}{7}\right)^{2}=\frac{20}{7}+\left(\frac{34}{7}\right)^{2}$$

Square \frac{34}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{68}{7}x+\frac{1156}{49}=\frac{20}{7}+\frac{1156}{49}$$

Add \frac{20}{7} to \frac{1156}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{68}{7}x+\frac{1156}{49}=\frac{1296}{49}$$

Factor x^{2}+\frac{68}{7}x+\frac{1156}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{34}{7}\right)^{2}=\frac{1296}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{34}{7}\right)^{2}}=\sqrt{\frac{1296}{49}}$$

Simplify.

$$x+\frac{34}{7}=\frac{36}{7} x+\frac{34}{7}=-\frac{36}{7}$$

Subtract \frac{34}{7} from both sides of the equation.

$$x=\frac{2}{7} x=-10$$