Solve 1/25(x-1)^2=1/9(x+1)^2

$$\frac { 1 } { 25 } ( x - 1 ) ^ { 2 } = \frac { 1 } { 9 } ( x + 1 ) ^ { 2 }$$

Solve for x

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

$$x=-4$$

Steps Using the Quadratic Formula

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

$$\frac{1}{25}\left(x^{2}-2x+1\right)=\frac{1}{9}\left(x+1\right)^{2}$$

Use the distributive property to multiply \frac{1}{25} by x^{2}-2x+1.

$$\frac{1}{25}x^{2}-\frac{2}{25}x+\frac{1}{25}=\frac{1}{9}\left(x+1\right)^{2}$$

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

$$\frac{1}{25}x^{2}-\frac{2}{25}x+\frac{1}{25}=\frac{1}{9}\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply \frac{1}{9} by x^{2}+2x+1.

$$\frac{1}{25}x^{2}-\frac{2}{25}x+\frac{1}{25}=\frac{1}{9}x^{2}+\frac{2}{9}x+\frac{1}{9}$$

Subtract \frac{1}{9}x^{2} from both sides.

$$\frac{1}{25}x^{2}-\frac{2}{25}x+\frac{1}{25}-\frac{1}{9}x^{2}=\frac{2}{9}x+\frac{1}{9}$$

Combine \frac{1}{25}x^{2} and -\frac{1}{9}x^{2} to get -\frac{16}{225}x^{2}.

$$-\frac{16}{225}x^{2}-\frac{2}{25}x+\frac{1}{25}=\frac{2}{9}x+\frac{1}{9}$$

Subtract \frac{2}{9}x from both sides.

$$-\frac{16}{225}x^{2}-\frac{2}{25}x+\frac{1}{25}-\frac{2}{9}x=\frac{1}{9}$$

Combine -\frac{2}{25}x and -\frac{2}{9}x to get -\frac{68}{225}x.

$$-\frac{16}{225}x^{2}-\frac{68}{225}x+\frac{1}{25}=\frac{1}{9}$$

Subtract \frac{1}{9} from both sides.

$$-\frac{16}{225}x^{2}-\frac{68}{225}x+\frac{1}{25}-\frac{1}{9}=0$$

Subtract \frac{1}{9} from \frac{1}{25} to get -\frac{16}{225}.

$$-\frac{16}{225}x^{2}-\frac{68}{225}x-\frac{16}{225}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{16}{225} for a, -\frac{68}{225} for b, and -\frac{16}{225} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{68}{225}\right)±\sqrt{\left(-\frac{68}{225}\right)^{2}-4\left(-\frac{16}{225}\right)\left(-\frac{16}{225}\right)}}{2\left(-\frac{16}{225}\right)}$$

Square -\frac{68}{225} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{68}{225}\right)±\sqrt{\frac{4624}{50625}-4\left(-\frac{16}{225}\right)\left(-\frac{16}{225}\right)}}{2\left(-\frac{16}{225}\right)}$$

Multiply -4 times -\frac{16}{225}.

$$x=\frac{-\left(-\frac{68}{225}\right)±\sqrt{\frac{4624}{50625}+\frac{64}{225}\left(-\frac{16}{225}\right)}}{2\left(-\frac{16}{225}\right)}$$

Multiply \frac{64}{225} times -\frac{16}{225} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{68}{225}\right)±\sqrt{\frac{4624-1024}{50625}}}{2\left(-\frac{16}{225}\right)}$$

Add \frac{4624}{50625} to -\frac{1024}{50625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{68}{225}\right)±\sqrt{\frac{16}{225}}}{2\left(-\frac{16}{225}\right)}$$

Take the square root of \frac{16}{225}.

$$x=\frac{-\left(-\frac{68}{225}\right)±\frac{4}{15}}{2\left(-\frac{16}{225}\right)}$$

The opposite of -\frac{68}{225} is \frac{68}{225}.

$$x=\frac{\frac{68}{225}±\frac{4}{15}}{2\left(-\frac{16}{225}\right)}$$

Multiply 2 times -\frac{16}{225}.

$$x=\frac{\frac{68}{225}±\frac{4}{15}}{-\frac{32}{225}}$$

Now solve the equation x=\frac{\frac{68}{225}±\frac{4}{15}}{-\frac{32}{225}} when ± is plus. Add \frac{68}{225} to \frac{4}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{128}{225}}{-\frac{32}{225}}$$

Divide \frac{128}{225} by -\frac{32}{225} by multiplying \frac{128}{225} by the reciprocal of -\frac{32}{225}.

$$x=-4$$

Now solve the equation x=\frac{\frac{68}{225}±\frac{4}{15}}{-\frac{32}{225}} when ± is minus. Subtract \frac{4}{15} from \frac{68}{225} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{8}{225}}{-\frac{32}{225}}$$

Divide \frac{8}{225} by -\frac{32}{225} by multiplying \frac{8}{225} by the reciprocal of -\frac{32}{225}.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$\frac{1}{25}\left(x^{2}-2x+1\right)=\frac{1}{9}\left(x+1\right)^{2}$$

Use the distributive property to multiply \frac{1}{25} by x^{2}-2x+1.

$$\frac{1}{25}x^{2}-\frac{2}{25}x+\frac{1}{25}=\frac{1}{9}\left(x+1\right)^{2}$$

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

$$\frac{1}{25}x^{2}-\frac{2}{25}x+\frac{1}{25}=\frac{1}{9}\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply \frac{1}{9} by x^{2}+2x+1.

$$\frac{1}{25}x^{2}-\frac{2}{25}x+\frac{1}{25}=\frac{1}{9}x^{2}+\frac{2}{9}x+\frac{1}{9}$$

Subtract \frac{1}{9}x^{2} from both sides.

$$\frac{1}{25}x^{2}-\frac{2}{25}x+\frac{1}{25}-\frac{1}{9}x^{2}=\frac{2}{9}x+\frac{1}{9}$$

Combine \frac{1}{25}x^{2} and -\frac{1}{9}x^{2} to get -\frac{16}{225}x^{2}.

$$-\frac{16}{225}x^{2}-\frac{2}{25}x+\frac{1}{25}=\frac{2}{9}x+\frac{1}{9}$$

Subtract \frac{2}{9}x from both sides.

$$-\frac{16}{225}x^{2}-\frac{2}{25}x+\frac{1}{25}-\frac{2}{9}x=\frac{1}{9}$$

Combine -\frac{2}{25}x and -\frac{2}{9}x to get -\frac{68}{225}x.

$$-\frac{16}{225}x^{2}-\frac{68}{225}x+\frac{1}{25}=\frac{1}{9}$$

Subtract \frac{1}{25} from both sides.

$$-\frac{16}{225}x^{2}-\frac{68}{225}x=\frac{1}{9}-\frac{1}{25}$$

Subtract \frac{1}{25} from \frac{1}{9} to get \frac{16}{225}.

$$-\frac{16}{225}x^{2}-\frac{68}{225}x=\frac{16}{225}$$

Divide both sides of the equation by -\frac{16}{225}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{16}{225}x^{2}-\frac{68}{225}x}{-\frac{16}{225}}=\frac{\frac{16}{225}}{-\frac{16}{225}}$$

Dividing by -\frac{16}{225} undoes the multiplication by -\frac{16}{225}.

$$x^{2}+\left(-\frac{\frac{68}{225}}{-\frac{16}{225}}\right)x=\frac{\frac{16}{225}}{-\frac{16}{225}}$$

Divide -\frac{68}{225} by -\frac{16}{225} by multiplying -\frac{68}{225} by the reciprocal of -\frac{16}{225}.

$$x^{2}+\frac{17}{4}x=\frac{\frac{16}{225}}{-\frac{16}{225}}$$

Divide \frac{16}{225} by -\frac{16}{225} by multiplying \frac{16}{225} by the reciprocal of -\frac{16}{225}.

$$x^{2}+\frac{17}{4}x=-1$$

Divide \frac{17}{4}, the coefficient of the x term, by 2 to get \frac{17}{8}. Then add the square of \frac{17}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{17}{4}x+\left(\frac{17}{8}\right)^{2}=-1+\left(\frac{17}{8}\right)^{2}$$

Square \frac{17}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{17}{4}x+\frac{289}{64}=-1+\frac{289}{64}$$

Add -1 to \frac{289}{64}.

$$x^{2}+\frac{17}{4}x+\frac{289}{64}=\frac{225}{64}$$

Factor x^{2}+\frac{17}{4}x+\frac{289}{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}.

$$\left(x+\frac{17}{8}\right)^{2}=\frac{225}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{17}{8}\right)^{2}}=\sqrt{\frac{225}{64}}$$

Simplify.

$$x+\frac{17}{8}=\frac{15}{8} x+\frac{17}{8}=-\frac{15}{8}$$

Subtract \frac{17}{8} from both sides of the equation.

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