Question

Solve the quadratic by completing the square x^2-4x-9=0

Answer 1

We are given the following quadratic equation

$x^2-4x-9=0$

When we have a quadratic equation of the form:

$x^2+bx+c=0$

Then, to complete the square we add and subtract the following expression:

$(\frac{b}{2})^2$

Replacing the value of "b"

$(\frac{4}{2})^2=2^2=4$

Adding and subtracting the term:

$x^2-4x+4-4-9$

Associating terms:

$(x^2-4x+4)+(-4-9)=0$

factoring the expression in the first parenthesis:

$(x-2)^2+(-4-9)=0$

Solving the operation in the second parenthesis:

$(x-2)^2-13=0$

Now we solve for "x", first by adding 13 to both sides:

$\begin{gathered} (x-2)^2-13+13=13 \\ (x-2)^2=13 \end{gathered}$

Now, we take square root on both sides:

$x-2=\sqrt[]{13}$

Now we add 2 to both sides:

$x=2\pm\sqrt[]{13}$

We have to possible values for "x", the first value is:

$x=2+\sqrt[]{13}=5.6$

The second value is:

$x=2-\sqrt[]{13}=-1.6$