Solve y^2-39y+324

$$y ^ { 2 } - 39 y + 324$$

Factor

$$\left(y-27\right)\left(y-12\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+324. To find a and b, set up a system to be solved.

$$a+b=-39 ab=1\times 324=324$$

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 324.

$$-1,-324 -2,-162 -3,-108 -4,-81 -6,-54 -9,-36 -12,-27 -18,-18$$

Calculate the sum for each pair.

$$-1-324=-325 -2-162=-164 -3-108=-111 -4-81=-85 -6-54=-60 -9-36=-45 -12-27=-39 -18-18=-36$$

The solution is the pair that gives sum -39.

$$a=-27 b=-12$$

Rewrite y^{2}-39y+324 as \left(y^{2}-27y\right)+\left(-12y+324\right).

$$\left(y^{2}-27y\right)+\left(-12y+324\right)$$

Factor out y in the first and -12 in the second group.

$$y\left(y-27\right)-12\left(y-27\right)$$

Factor out common term y-27 by using distributive property.

$$\left(y-27\right)\left(y-12\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$y^{2}-39y+324=0$$

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.

$$y=\frac{-\left(-39\right)±\sqrt{\left(-39\right)^{2}-4\times 324}}{2}$$

Square -39.

$$y=\frac{-\left(-39\right)±\sqrt{1521-4\times 324}}{2}$$

Multiply -4 times 324.

$$y=\frac{-\left(-39\right)±\sqrt{1521-1296}}{2}$$

Add 1521 to -1296.

$$y=\frac{-\left(-39\right)±\sqrt{225}}{2}$$

Take the square root of 225.

$$y=\frac{-\left(-39\right)±15}{2}$$

The opposite of -39 is 39.

$$y=\frac{39±15}{2}$$

Now solve the equation y=\frac{39±15}{2} when ± is plus. Add 39 to 15.

$$y=\frac{54}{2}$$

Divide 54 by 2.

$$y=27$$

Now solve the equation y=\frac{39±15}{2} when ± is minus. Subtract 15 from 39.

$$y=\frac{24}{2}$$

Divide 24 by 2.

$$y=12$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 27 for x_{1} and 12 for x_{2}.

$$y^{2}-39y+324=\left(y-27\right)\left(y-12\right)$$

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.

$$x ^ 2 -39x +324 = 0$$

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

$$r + s = 39 rs = 324$$

Two numbers r and s sum up to 39 exactly when the average of the two numbers is \frac{1}{2}*39 = \frac{39}{2}. 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. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{39}{2} - u s = \frac{39}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 324

$$(\frac{39}{2} - u) (\frac{39}{2} + u) = 324$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{1521}{4} - u^2 = 324$$

Simplify the expression by subtracting \frac{1521}{4} on both sides

$$-u^2 = 324-\frac{1521}{4} = -\frac{225}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{225}{4} u = \pm\sqrt{\frac{225}{4}} = \pm \frac{15}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{39}{2} - \frac{15}{2} = 12 s = \frac{39}{2} + \frac{15}{2} = 27$$

Evaluate

$$\left(y-27\right)\left(y-12\right)$$