# Question #3c21d

Aug 3, 2016

In line $3$.

#### Explanation:

Let me start by rewriting the five lines

$\text{L1: } 4 {s}^{2} - 40 s + 100 = 0$

$\text{L2: } {s}^{2} - 10 s + 25 = 0$

$\text{L3: } s = \frac{- 10 \pm \sqrt{{10}^{2} - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

$\text{L4: } s = \frac{- 10 \pm 0}{2}$

$\text{L5: } s = - \frac{10}{2}$

$\text{L6: } s = - 5$

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

The way I see it, the error occurs in line $3$, $\text{L3}$. Here's why.

$4 {s}^{2} - 40 s + 100 = 0$

To get to $\text{L2}$, you need to divide all the terms by $4$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} {s}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} - \frac{40 s}{4} + \frac{100}{4} = 0$

${s}^{2} - 10 s + 25 = 0$

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

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

takes the form

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\left\{\begin{matrix}a = 1 \\ b = - 10 \\ c = 25\end{matrix}\right.$

This means that line $3$ should read

${s}_{1 , 2} = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

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

$\text{L3: } {s}_{1 , 2} = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \cdot 1 \cdot 25}}{2 \cdot 1}$

$\text{L4: } {s}_{1 , 2} = \frac{10 \pm 0}{2}$

$\text{L5: } s = \frac{10}{2}$

$\text{L6: } s = 5$