Question #3c21d

1 Answer
Aug 3, 2016

Answer:

In line #3#.

Explanation:

Let me start by rewriting the five lines

#"L1: " 4s^2 - 40s + 100 = 0#

#"L2: " s^2 - 10s + 25 = 0#

#"L3: " s = (-10 +- sqrt( 10^2 - 4 * 1 * 25))/(2 * 1)#

#"L4: " s = (-10 +- 0)/2#

#"L5: " s = -10/2#

#"L6: " s = -5#

#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

The way I see it, the error occurs in line #3#, #"L3"#. Here's why.

Your starting quadratic equation looks like this

#4s^2 - 40s + 100 = 0#

To get to #"L2"#, you need to divide all the terms by #4#

#(color(red)(cancel(color(black)(4)))s^2)/color(red)(cancel(color(black)(4))) - (40s)/4 + 100/4 = 0#

#s^2 - 10s + 25 = 0#

So far, so good. Now comes the interesting part. You can get to line #3# by using the quadratic formula, which for a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

takes the form

#color(blue)(|bar(ul(color(white)(a/a)x_(1,2) = (-b +- sqrt(b^2 - 4 * a * c))/(2 * a) color(white)(a/a)|)))#

For your quadratic, you have

#{(a = 1), (b = -10), (c = 25) :}#

This means that line #3# should read

#s_(1,2) = (- (-10) +- sqrt( (-10)^2 - 4 * 1 * 25))/(2 * 1)#

As you can see, your example uses #b = 10# instead of #b = -10#. This represents the first mistake made when trying to solve this quadratic. From this point on, all the remaining lines are incorrect.

Therefore, you should have

#"L3: " s_(1,2) = (- (-10) +- sqrt( (-10)^2 - 4 * 1 * 25))/(2 * 1)#

#"L4: " s_(1,2) = (10 +- 0)/2#

#"L5: " s = 10/2#

#"L6: " s = 5#