Solve (x-3)^2=x-1

$$( x - 3 ) ^ { 2 } = x - 1$$

Solve for x

$$x=2$$

$$x=5$$

Steps Using Factoring

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

$$x^{2}-6x+9=x-1$$

Subtract x from both sides.

$$x^{2}-6x+9-x=-1$$

Combine -6x and -x to get -7x.

$$x^{2}-7x+9=-1$$

Add 1 to both sides.

$$x^{2}-7x+9+1=0$$

Add 9 and 1 to get 10.

$$x^{2}-7x+10=0$$

To solve the equation, factor x^{2}-7x+10 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-7 ab=10$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-5\right)\left(x-2\right)$$

To find equation solutions, solve x-5=0 and x-2=0.

$$x=5 x=2$$

Steps Using Factoring By Grouping

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

$$x^{2}-6x+9=x-1$$

Subtract x from both sides.

$$x^{2}-6x+9-x=-1$$

Combine -6x and -x to get -7x.

$$x^{2}-7x+9=-1$$

Add 1 to both sides.

$$x^{2}-7x+9+1=0$$

Add 9 and 1 to get 10.

$$x^{2}-7x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 10=10$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite x^{2}-7x+10 as \left(x^{2}-5x\right)+\left(-2x+10\right).

$$\left(x^{2}-5x\right)+\left(-2x+10\right)$$

Factor out x in the first and -2 in the second group.

$$x\left(x-5\right)-2\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-2\right)$$

To find equation solutions, solve x-5=0 and x-2=0.

$$x=5 x=2$$

Steps Using the Quadratic Formula

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

$$x^{2}-6x+9=x-1$$

Subtract x from both sides.

$$x^{2}-6x+9-x=-1$$

Combine -6x and -x to get -7x.

$$x^{2}-7x+9=-1$$

Add 1 to both sides.

$$x^{2}-7x+9+1=0$$

Add 9 and 1 to get 10.

$$x^{2}-7x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 10}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 10}}{2}$$

Multiply -4 times 10.

$$x=\frac{-\left(-7\right)±\sqrt{49-40}}{2}$$

Add 49 to -40.

$$x=\frac{-\left(-7\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$x=\frac{-\left(-7\right)±3}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±3}{2}$$

Now solve the equation x=\frac{7±3}{2} when ± is plus. Add 7 to 3.

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

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{7±3}{2} when ± is minus. Subtract 3 from 7.

$$x=\frac{4}{2}$$

Divide 4 by 2.

$$x=2$$

The equation is now solved.

$$x=5 x=2$$

Steps for Completing the Square

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

$$x^{2}-6x+9=x-1$$

Subtract x from both sides.

$$x^{2}-6x+9-x=-1$$

Combine -6x and -x to get -7x.

$$x^{2}-7x+9=-1$$

Subtract 9 from both sides.

$$x^{2}-7x=-1-9$$

Subtract 9 from -1 to get -10.

$$x^{2}-7x=-10$$

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

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-10+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=-10+\frac{49}{4}$$

Add -10 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{9}{4}$$

Factor x^{2}-7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{3}{2} x-\frac{7}{2}=-\frac{3}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=5 x=2$$