Solve r^2-5r+9=r

$$r ^ { 2 } - 5 r + 9 = r$$

Solve for r

$$r=3$$

Steps Using Factoring

Subtract r from both sides.

$$r^{2}-5r+9-r=0$$

Combine -5r and -r to get -6r.

$$r^{2}-6r+9=0$$

To solve the equation, factor r^{2}-6r+9 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -6.

$$a=-3 b=-3$$

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

$$\left(r-3\right)\left(r-3\right)$$

Rewrite as a binomial square.

$$\left(r-3\right)^{2}$$

To find equation solution, solve r-3=0.

$$r=3$$

Steps Using Factoring By Grouping

Subtract r from both sides.

$$r^{2}-5r+9-r=0$$

Combine -5r and -r to get -6r.

$$r^{2}-6r+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as r^{2}+ar+br+9. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 9=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -6.

$$a=-3 b=-3$$

Rewrite r^{2}-6r+9 as \left(r^{2}-3r\right)+\left(-3r+9\right).

$$\left(r^{2}-3r\right)+\left(-3r+9\right)$$

Factor out r in the first and -3 in the second group.

$$r\left(r-3\right)-3\left(r-3\right)$$

Factor out common term r-3 by using distributive property.

$$\left(r-3\right)\left(r-3\right)$$

Rewrite as a binomial square.

$$\left(r-3\right)^{2}$$

To find equation solution, solve r-3=0.

$$r=3$$

Steps Using the Quadratic Formula

Subtract r from both sides.

$$r^{2}-5r+9-r=0$$

Combine -5r and -r to get -6r.

$$r^{2}-6r+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -6 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9}}{2}$$

Square -6.

$$r=\frac{-\left(-6\right)±\sqrt{36-4\times 9}}{2}$$

Multiply -4 times 9.

$$r=\frac{-\left(-6\right)±\sqrt{36-36}}{2}$$

Add 36 to -36.

$$r=\frac{-\left(-6\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$r=-\frac{-6}{2}$$

The opposite of -6 is 6.

$$r=\frac{6}{2}$$

Divide 6 by 2.

$$r=3$$

Steps for Completing the Square

Subtract r from both sides.

$$r^{2}-5r+9-r=0$$

Combine -5r and -r to get -6r.

$$r^{2}-6r+9=0$$

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

$$\left(r-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$r-3=0 r-3=0$$

Add 3 to both sides of the equation.

$$r=3 r=3$$

The equation is now solved. Solutions are the same.

$$r=3$$