Solve L^2-100L+1600^circ

$$L ^ { 2 } - 100 L + 1600 ^ { \circ }$$

Factor

$$\left(L-80\right)\left(L-20\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as L^{2}+aL+bL+1600. To find a and b, set up a system to be solved.

$$a+b=-100 ab=1\times 1600=1600$$

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 1600.

$$-1,-1600 -2,-800 -4,-400 -5,-320 -8,-200 -10,-160 -16,-100 -20,-80 -25,-64 -32,-50 -40,-40$$

Calculate the sum for each pair.

$$-1-1600=-1601 -2-800=-802 -4-400=-404 -5-320=-325 -8-200=-208 -10-160=-170 -16-100=-116 -20-80=-100 -25-64=-89 -32-50=-82 -40-40=-80$$

The solution is the pair that gives sum -100.

$$a=-80 b=-20$$

Rewrite L^{2}-100L+1600 as \left(L^{2}-80L\right)+\left(-20L+1600\right).

$$\left(L^{2}-80L\right)+\left(-20L+1600\right)$$

Factor out L in the first and -20 in the second group.

$$L\left(L-80\right)-20\left(L-80\right)$$

Factor out common term L-80 by using distributive property.

$$\left(L-80\right)\left(L-20\right)$$

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.

$$L^{2}-100L+1600=0$$

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.

$$L=\frac{-\left(-100\right)±\sqrt{\left(-100\right)^{2}-4\times 1600}}{2}$$

Square -100.

$$L=\frac{-\left(-100\right)±\sqrt{10000-4\times 1600}}{2}$$

Multiply -4 times 1600.

$$L=\frac{-\left(-100\right)±\sqrt{10000-6400}}{2}$$

Add 10000 to -6400.

$$L=\frac{-\left(-100\right)±\sqrt{3600}}{2}$$

Take the square root of 3600.

$$L=\frac{-\left(-100\right)±60}{2}$$

The opposite of -100 is 100.

$$L=\frac{100±60}{2}$$

Now solve the equation L=\frac{100±60}{2} when ± is plus. Add 100 to 60.

$$L=\frac{160}{2}$$

Divide 160 by 2.

$$L=80$$

Now solve the equation L=\frac{100±60}{2} when ± is minus. Subtract 60 from 100.

$$L=\frac{40}{2}$$

Divide 40 by 2.

$$L=20$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 80 for x_{1} and 20 for x_{2}.

$$L^{2}-100L+1600=\left(L-80\right)\left(L-20\right)$$

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.

$$x ^ 2 -100x +1600 = 0$$

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

$$r + s = 100 rs = 1600$$

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 = 1600

$$(50 - u) (50 + u) = 1600$$

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

$$2500 - u^2 = 1600$$

Simplify the expression by subtracting 2500 on both sides

$$-u^2 = 1600-2500 = -900$$

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

$$u^2 = 900 u = \pm\sqrt{900} = \pm 30$$

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

$$r =50 - 30 = 20 s = 50 + 30 = 80$$

Evaluate

$$\left(L-80\right)\left(L-20\right)$$