To solve the equation, factor $x^{2}-6x+9$ 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.
$$a+b=-6$$ $$ab=9$$
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 $\left(x+a\right)\left(x+b\right)$ using the obtained values.
$$\left(x-3\right)\left(x-3\right)$$
Rewrite as a binomial square.
$$\left(x-3\right)^{2}$$
To find equation solution, solve $x-3=0$.
$$x=3$$
Steps Using Factoring By Grouping
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx+9$. 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 $x^{2}-6x+9$ as $\left(x^{2}-3x\right)+\left(-3x+9\right)$.
$$\left(x^{2}-3x\right)+\left(-3x+9\right)$$
Factor out $x$ in the first and $-3$ in the second group.
$$x\left(x-3\right)-3\left(x-3\right)$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(x-3\right)$$
Rewrite as a binomial square.
$$\left(x-3\right)^{2}$$
To find equation solution, solve $x-3=0$.
$$x=3$$
Steps Using the Quadratic Formula
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$x^{2}-6x+9=0$$
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, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
$$x^{2}-6x+9=0$$
Factor $x^{2}-6x+9$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(x-3\right)^{2}=0$$
Take the square root of both sides of the equation.
$$\sqrt{\left(x-3\right)^{2}}=\sqrt{0}$$
Simplify.
$$x-3=0$$ $$x-3=0$$
Add $3$ to both sides of the equation.
$$x=3$$ $$x=3$$
The equation is now solved. Solutions are the same.
$$x=3$$
Steps Using Direct Factoring Method
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.
$$x ^ 2 -6x +9 = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = 6 $$ $$ rs = 9$$
Two numbers $r$ and $s$ sum up to $6$ exactly when the average of the two numbers is $\frac{1}{2}*6 = 3$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = 3 - u$$ $$s = 3 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 9$
$$(3 - u) (3 + u) = 9$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$9 - u^2 = 9$$
Simplify the expression by subtracting $9$ on both sides
$$-u^2 = 9-9 = 0$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
$$u^2 = 0$$ $$u = 0 $$
The factors $r$ and $s$ are the solutions to the quadratic equation. Substitute the value of $u$ to compute the $r$ and $s$.
