Solve b^2-100b+2475

$$b ^ { 2 } - 100 b + 2475$$

Factor

$$\left(b-55\right)\left(b-45\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as b^{2}+pb+qb+2475. To find p and q, set up a system to be solved.

$$p+q=-100 pq=1\times 2475=2475$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 2475.

$$-1,-2475 -3,-825 -5,-495 -9,-275 -11,-225 -15,-165 -25,-99 -33,-75 -45,-55$$

Calculate the sum for each pair.

$$-1-2475=-2476 -3-825=-828 -5-495=-500 -9-275=-284 -11-225=-236 -15-165=-180 -25-99=-124 -33-75=-108 -45-55=-100$$

The solution is the pair that gives sum -100.

$$p=-55 q=-45$$

Rewrite b^{2}-100b+2475 as \left(b^{2}-55b\right)+\left(-45b+2475\right).

$$\left(b^{2}-55b\right)+\left(-45b+2475\right)$$

Factor out b in the first and -45 in the second group.

$$b\left(b-55\right)-45\left(b-55\right)$$

Factor out common term b-55 by using distributive property.

$$\left(b-55\right)\left(b-45\right)$$

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.

$$b^{2}-100b+2475=0$$

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.

$$b=\frac{-\left(-100\right)±\sqrt{\left(-100\right)^{2}-4\times 2475}}{2}$$

Square -100.

$$b=\frac{-\left(-100\right)±\sqrt{10000-4\times 2475}}{2}$$

Multiply -4 times 2475.

$$b=\frac{-\left(-100\right)±\sqrt{10000-9900}}{2}$$

Add 10000 to -9900.

$$b=\frac{-\left(-100\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$b=\frac{-\left(-100\right)±10}{2}$$

The opposite of -100 is 100.

$$b=\frac{100±10}{2}$$

Now solve the equation b=\frac{100±10}{2} when ± is plus. Add 100 to 10.

$$b=\frac{110}{2}$$

Divide 110 by 2.

$$b=55$$

Now solve the equation b=\frac{100±10}{2} when ± is minus. Subtract 10 from 100.

$$b=\frac{90}{2}$$

Divide 90 by 2.

$$b=45$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 55 for x_{1} and 45 for x_{2}.

$$b^{2}-100b+2475=\left(b-55\right)\left(b-45\right)$$

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.

$$x ^ 2 -100x +2475 = 0$$

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

$$r + s = 100 rs = 2475$$

Two numbers r and s sum up to 100 exactly when the average of the two numbers is \frac{1}{2}*100 = 50. 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>

$$r = 50 - u s = 50 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 2475

$$(50 - u) (50 + u) = 2475$$

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

$$2500 - u^2 = 2475$$

Simplify the expression by subtracting 2500 on both sides

$$-u^2 = 2475-2500 = -25$$

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

$$u^2 = 25 u = \pm\sqrt{25} = \pm 5$$

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

$$r =50 - 5 = 45 s = 50 + 5 = 55$$

Evaluate

$$\left(b-55\right)\left(b-45\right)$$