+1252 We can factor this quadratic!
\((x-5)(x-5)\) so only 5 fits as \(x\) . The answer is B.
You are very welcome!
:P
+4622 I'm going to take you through three steps on how to solve this!
1. Quadratic Formula
The quadratic formula, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) is a very neat formula on how to find the roots! Plugging the values into the equation, we yield \(x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\cdot \:25}}{2\cdot \:1}\) . Solving and doing all the calculations, we get \(\frac{-\left(-10\right)}{2\cdot \:1}=\boxed{5}\), and that is our only solution!
2. Factoring
We can factor \(x^2-10x+25=0\) into \(\left(x-5\right)^2\) , after trying! Now, set \(x-5=0\) , and our only solution is \(\boxed{5}.\)
3. Completing the Square
To complete the square, we first subtract \(25\) from both sides, leaving us with \(x^2-10x=-25\). Our goal is to write it in the form \(x^2+2ax+a^2=\left(x+a\right)^2\) , so we solve for \(a\) , which is \(-5.\) We add this to get \(x^2-10x+\left(-5\right)^2=-25+\left(-5\right)^2\) . Solving, we get , \(\left(x-5\right)^2=0\)
and our answer is \(\boxed{x=5}\).
Too many steps! The answer is (B).
