+0

# Help 5 ​

+1
127
2
+4684

Help 5

Nov 23, 2018

#1
+626
0

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

You are very welcome!

:P

Nov 23, 2018
#2
+4078
+2

I'm going to take you through three steps on how to solve this!

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