Respuesta :
If i'm correct your answer should be x = [tex]3 \sqrt{10} + 6[/tex]. Hope it helps!
Answer:
[tex]x=6-3\sqrt{10}\text{ or }x=6+3\sqrt{10}[/tex]
Step-by-step explanation:
We have been given an equation [tex]x^2-12x+36=90[/tex]. We are asked to solve for x in our given equation.
We can see that left side of our given equation is a perfect square.
[tex](x-6)^2=90[/tex]
We will take square root of both sides of our given equation to solve for x.
[tex]\sqrt{(x-6)^2}=\pm\sqrt{90}[/tex]
Using radical rule [tex]\sqrt[n]{a^n}=a[/tex], we will get:
[tex]x-6=\pm\sqrt{90}[/tex]
[tex]x-6=\pm\sqrt{9\cdot 10}[/tex]
[tex]x-6=\pm\sqrt{3^2\cdot 10}[/tex]
[tex]x-6=\pm\sqrt{3^2}\cdot \sqrt{10}[/tex]
Using radical rule [tex]\sqrt[n]{a^n}=a[/tex], we will get:
[tex]x-6=\pm3\sqrt{10}[/tex]
Now, we will add 6 on both sides of our given equation as:
[tex]x-6+6=\pm 3\sqrt{10}+6[/tex]
[tex]x=\pm 3\sqrt{10}+6[/tex]
[tex]x=-3\sqrt{10}+6\text{ or }x=3\sqrt{10}+6[/tex]
[tex]x=6-3\sqrt{10}\text{ or }x=6+3\sqrt{10}[/tex]
Therefore, the solution for our given equation would be [tex]x=6-3\sqrt{10}\text{ or }x=6+3\sqrt{10}[/tex].
