To solve the equation, factor $w^{2}-12w+36$ using formula $w^{2}+\left(a+b\right)w+ab=\left(w+a\right)\left(w+b\right)$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-12$$ $$ab=36$$
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 $36$.
Rewrite factored expression $\left(w+a\right)\left(w+b\right)$ using the obtained values.
$$\left(w-6\right)\left(w-6\right)$$
Rewrite as a binomial square.
$$\left(w-6\right)^{2}$$
To find equation solution, solve $w-6=0$.
$$w=6$$
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 $w^{2}+aw+bw+36$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-12$$ $$ab=1\times 36=36$$
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 $36$.
Rewrite $w^{2}-12w+36$ as $\left(w^{2}-6w\right)+\left(-6w+36\right)$.
$$\left(w^{2}-6w\right)+\left(-6w+36\right)$$
Factor out $w$ in the first and $-6$ in the second group.
$$w\left(w-6\right)-6\left(w-6\right)$$
Factor out common term $w-6$ by using distributive property.
$$\left(w-6\right)\left(w-6\right)$$
Rewrite as a binomial square.
$$\left(w-6\right)^{2}$$
To find equation solution, solve $w-6=0$.
$$w=6$$
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.
$$w^{2}-12w+36=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-12$ for $b$, and $36$ 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 $x^{2}+bx=c$.
$$w^{2}-12w+36=0$$
Factor $w^{2}-12w+36$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(w-6\right)^{2}=0$$
Take the square root of both sides of the equation.
$$\sqrt{\left(w-6\right)^{2}}=\sqrt{0}$$
Simplify.
$$w-6=0$$ $$w-6=0$$
Add $6$ to both sides of the equation.
$$w=6$$ $$w=6$$
The equation is now solved. Solutions are the same.
$$w=6$$
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 $x^2+Bx+C=0$.
$$x ^ 2 -12x +36 = 0$$
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 $rs = C$
$$r + s = 12 $$ $$ rs = 36$$
Two numbers $r$ and $s$ sum up to $12$ exactly when the average of the two numbers is $\frac{1}{2}*12 = 6$. 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 $u$.
$$r = 6 - u$$ $$s = 6 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 36$
$$(6 - u) (6 + u) = 36$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$36 - u^2 = 36$$
Simplify the expression by subtracting $36$ on both sides
$$-u^2 = 36-36 = 0$$
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 $s$.
