Question

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

Answer 1

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+((1)/(2)\cdot(-2))^2=38+((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}`