Solve left(x-1right)^2+left(8x-8right)^2=4

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

Solve for x

$$x=\frac{2\sqrt{65}}{65}+1\approx 1.248069469$$

$$x=-\frac{2\sqrt{65}}{65}+1\approx 0.751930531$$

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}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8x-8\right)^{2}.

$$x^{2}-2x+1+64x^{2}-128x+64=4$$

Combine x^{2} and 64x^{2} to get 65x^{2}.

$$65x^{2}-2x+1-128x+64=4$$

Combine -2x and -128x to get -130x.

$$65x^{2}-130x+1+64=4$$

Add 1 and 64 to get 65.

$$65x^{2}-130x+65=4$$

Subtract 4 from both sides.

$$65x^{2}-130x+65-4=0$$

Subtract 4 from 65 to get 61.

$$65x^{2}-130x+61=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 65 for a, -130 for b, and 61 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-130\right)±\sqrt{\left(-130\right)^{2}-4\times 65\times 61}}{2\times 65}$$

Square -130.

$$x=\frac{-\left(-130\right)±\sqrt{16900-4\times 65\times 61}}{2\times 65}$$

Multiply -4 times 65.

$$x=\frac{-\left(-130\right)±\sqrt{16900-260\times 61}}{2\times 65}$$

Multiply -260 times 61.

$$x=\frac{-\left(-130\right)±\sqrt{16900-15860}}{2\times 65}$$

Add 16900 to -15860.

$$x=\frac{-\left(-130\right)±\sqrt{1040}}{2\times 65}$$

Take the square root of 1040.

$$x=\frac{-\left(-130\right)±4\sqrt{65}}{2\times 65}$$

The opposite of -130 is 130.

$$x=\frac{130±4\sqrt{65}}{2\times 65}$$

Multiply 2 times 65.

$$x=\frac{130±4\sqrt{65}}{130}$$

Now solve the equation x=\frac{130±4\sqrt{65}}{130} when ± is plus. Add 130 to 4\sqrt{65}.

$$x=\frac{4\sqrt{65}+130}{130}$$

Divide 130+4\sqrt{65} by 130.

$$x=\frac{2\sqrt{65}}{65}+1$$

Now solve the equation x=\frac{130±4\sqrt{65}}{130} when ± is minus. Subtract 4\sqrt{65} from 130.

$$x=\frac{130-4\sqrt{65}}{130}$$

Divide 130-4\sqrt{65} by 130.

$$x=-\frac{2\sqrt{65}}{65}+1$$

The equation is now solved.

$$x=\frac{2\sqrt{65}}{65}+1 x=-\frac{2\sqrt{65}}{65}+1$$

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}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8x-8\right)^{2}.

$$x^{2}-2x+1+64x^{2}-128x+64=4$$

Combine x^{2} and 64x^{2} to get 65x^{2}.

$$65x^{2}-2x+1-128x+64=4$$

Combine -2x and -128x to get -130x.

$$65x^{2}-130x+1+64=4$$

Add 1 and 64 to get 65.

$$65x^{2}-130x+65=4$$

Subtract 65 from both sides.

$$65x^{2}-130x=4-65$$

Subtract 65 from 4 to get -61.

$$65x^{2}-130x=-61$$

Divide both sides by 65.

$$\frac{65x^{2}-130x}{65}=-\frac{61}{65}$$

Dividing by 65 undoes the multiplication by 65.

$$x^{2}+\left(-\frac{130}{65}\right)x=-\frac{61}{65}$$

Divide -130 by 65.

$$x^{2}-2x=-\frac{61}{65}$$

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.

$$x^{2}-2x+1=-\frac{61}{65}+1$$

Add -\frac{61}{65} to 1.

$$x^{2}-2x+1=\frac{4}{65}$$

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}.

$$\left(x-1\right)^{2}=\frac{4}{65}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{\frac{4}{65}}$$

Simplify.

$$x-1=\frac{2\sqrt{65}}{65} x-1=-\frac{2\sqrt{65}}{65}$$

Add 1 to both sides of the equation.

$$x=\frac{2\sqrt{65}}{65}+1 x=-\frac{2\sqrt{65}}{65}+1$$