Question

Solve the quadratic equation by completing the square.x^2-10x+23=0First Tuesday appropriate form and fill in the blanks with the correct numbers. Then solve the equation. If there’s more than one solution separate them with commas.

Answer 1

Form part

We have the following quadratic equation:

$x^2-10x+23=0$

which can be rewritten as

$x^2-2(5x)+23=0$

By comparing with the perfect square binomial:

$\begin{gathered} (x-a)^2=x^2-2ax+a^2 \\ \text{then} \\ (x-a)-a^2=x^2-2ax \end{gathered}$

we can see that a is 5. Then, we have that

$x^2-10x+23=(x-5)^2-5^2+23$

which gives

$\begin{gathered} x^2-10x+23=(x-5)^2-25+23 \\ or,\text{ equivalently,} \\ x^2-10x+23=(x-5)^2-2 \end{gathered}$

Then, the answer of the form part is

$(x-5)^2-2$

Solution part

Now, lets solve the equation

$(x-5)^2-2=0$

By moving -2 to the right hand side, we have

$(x-5)^2=2$

By applying square root in both side, we get

$x-5=\pm\sqrt[]{2}$

then, we have 2 solutions:

Solution 1:

$x=5+\sqrt[]{2}$

Solution 2:

$x=5-\sqrt[]{2}$