Solve y^2-45y+506

$$y ^ { 2 } - 45 y + 506$$

Factor

$$\left(y-23\right)\left(y-22\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+506. To find a and b, set up a system to be solved.

$$a+b=-45 ab=1\times 506=506$$

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 506.

$$-1,-506 -2,-253 -11,-46 -22,-23$$

Calculate the sum for each pair.

$$-1-506=-507 -2-253=-255 -11-46=-57 -22-23=-45$$

The solution is the pair that gives sum -45.

$$a=-23 b=-22$$

Rewrite y^{2}-45y+506 as \left(y^{2}-23y\right)+\left(-22y+506\right).

$$\left(y^{2}-23y\right)+\left(-22y+506\right)$$

Factor out y in the first and -22 in the second group.

$$y\left(y-23\right)-22\left(y-23\right)$$

Factor out common term y-23 by using distributive property.

$$\left(y-23\right)\left(y-22\right)$$

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.

$$y^{2}-45y+506=0$$

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.

$$y=\frac{-\left(-45\right)±\sqrt{\left(-45\right)^{2}-4\times 506}}{2}$$

Square -45.

$$y=\frac{-\left(-45\right)±\sqrt{2025-4\times 506}}{2}$$

Multiply -4 times 506.

$$y=\frac{-\left(-45\right)±\sqrt{2025-2024}}{2}$$

Add 2025 to -2024.

$$y=\frac{-\left(-45\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$y=\frac{-\left(-45\right)±1}{2}$$

The opposite of -45 is 45.

$$y=\frac{45±1}{2}$$

Now solve the equation y=\frac{45±1}{2} when ± is plus. Add 45 to 1.

$$y=\frac{46}{2}$$

Divide 46 by 2.

$$y=23$$

Now solve the equation y=\frac{45±1}{2} when ± is minus. Subtract 1 from 45.

$$y=\frac{44}{2}$$

Divide 44 by 2.

$$y=22$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 23 for x_{1} and 22 for x_{2}.

$$y^{2}-45y+506=\left(y-23\right)\left(y-22\right)$$

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.

$$x ^ 2 -45x +506 = 0$$

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

$$r + s = 45 rs = 506$$

Two numbers r and s sum up to 45 exactly when the average of the two numbers is \frac{1}{2}*45 = \frac{45}{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>

$$r = \frac{45}{2} - u s = \frac{45}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 506

$$(\frac{45}{2} - u) (\frac{45}{2} + u) = 506$$

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

$$\frac{2025}{4} - u^2 = 506$$

Simplify the expression by subtracting \frac{2025}{4} on both sides

$$-u^2 = 506-\frac{2025}{4} = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =\frac{45}{2} - \frac{1}{2} = 22 s = \frac{45}{2} + \frac{1}{2} = 23$$

Evaluate

$$\left(y-23\right)\left(y-22\right)$$