Solve (x-3/5)^2=2/5+9/25

$$( x - \frac { 3 } { 5 } ) ^ { 2 } = \frac { 2 } { 5 } + \frac { 9 } { 25 }$$

Solve for x

$$x = \frac{\sqrt{19} + 3}{5} \approx 1.471779789$$

$$x=\frac{3-\sqrt{19}}{5}\approx -0.271779789$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{3}{5}\right)^{2}.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{2}{5}+\frac{9}{25}$$

Add \frac{2}{5} and \frac{9}{25} to get \frac{19}{25}.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{19}{25}$$

Subtract \frac{19}{25} from both sides.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}-\frac{19}{25}=0$$

Subtract \frac{19}{25} from \frac{9}{25} to get -\frac{2}{5}.

$$x^{2}-\frac{6}{5}x-\frac{2}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -\frac{6}{5} for b, and -\frac{2}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\left(-\frac{6}{5}\right)^{2}-4\left(-\frac{2}{5}\right)}}{2}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{36}{25}-4\left(-\frac{2}{5}\right)}}{2}$$

Multiply -4 times -\frac{2}{5}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{36}{25}+\frac{8}{5}}}{2}$$

Add \frac{36}{25} to \frac{8}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{6}{5}\right)±\sqrt{\frac{76}{25}}}{2}$$

Take the square root of \frac{76}{25}.

$$x=\frac{-\left(-\frac{6}{5}\right)±\frac{2\sqrt{19}}{5}}{2}$$

The opposite of -\frac{6}{5} is \frac{6}{5}.

$$x=\frac{\frac{6}{5}±\frac{2\sqrt{19}}{5}}{2}$$

Now solve the equation x=\frac{\frac{6}{5}±\frac{2\sqrt{19}}{5}}{2} when ± is plus. Add \frac{6}{5} to \frac{2\sqrt{19}}{5}.

$$x=\frac{2\sqrt{19}+6}{2\times 5}$$

Divide \frac{6+2\sqrt{19}}{5} by 2.

$$x=\frac{\sqrt{19}+3}{5}$$

Now solve the equation x=\frac{\frac{6}{5}±\frac{2\sqrt{19}}{5}}{2} when ± is minus. Subtract \frac{2\sqrt{19}}{5} from \frac{6}{5}.

$$x=\frac{6-2\sqrt{19}}{2\times 5}$$

Divide \frac{6-2\sqrt{19}}{5} by 2.

$$x=\frac{3-\sqrt{19}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{19}+3}{5} x=\frac{3-\sqrt{19}}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{3}{5}\right)^{2}.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{2}{5}+\frac{9}{25}$$

Add \frac{2}{5} to \frac{9}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{6}{5}x+\frac{9}{25}=\frac{19}{25}$$

Factor x^{2}-\frac{6}{5}x+\frac{9}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{5}\right)^{2}=\frac{19}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{5}\right)^{2}}=\sqrt{\frac{19}{25}}$$

Simplify.

$$x-\frac{3}{5}=\frac{\sqrt{19}}{5} x-\frac{3}{5}=-\frac{\sqrt{19}}{5}$$

Add \frac{3}{5} to both sides of the equation.

$$x=\frac{\sqrt{19}+3}{5} x=\frac{3-\sqrt{19}}{5}$$