Question

Hi, can you help to find (all the roots/zeros), please!!!

Answer 1

Solution

Given the quadratic equation

$x^2-2x-38=0$

we need to find the zeros of the equation

To do that, we use the completing the square method

Step 1. Add 38 to both sides

$\begin{gathered} \Rightarrow x^2-2x-38+38=38 \\ \\ \Rightarrow x^2-2x=38 \end{gathered}$

Step 2: add the square of half of the coefficient of x to both sides

That is;

$\begin{gathered} \Rightarrow x^2-2x+(\frac{1}{2}\cdot(-2))^2=38+(\frac{1}{2}\cdot(-2))^2 \\ \\ \Rightarrow x^2-2x+1=38+1 \\ \\ \Rightarrow x^2-2x+1=39 \\ \\ \Rightarrow(x-1)^2=39 \end{gathered}$

Step 3: Simplify the above expression;

$\begin{gathered} \Rightarrow x-1=\pm\sqrt[]{39} \\ \\ \Rightarrow x=1\pm\sqrt[]{39} \end{gathered}$