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Help 5

 Nov 23, 2018
 #1
avatar+626 
0

We can factor this quadratic!

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

 

You are very welcome!

:P

 Nov 23, 2018
 #2
avatar+4078 
+2

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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