Help 5

 Nov 23, 2018

We can factor this quadratic!

\((x-5)(x-5)\) so only 5 fits as \(x\) . The answer is B.


You are very welcome!


 Nov 23, 2018

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).

 Nov 23, 2018

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