square number / perfekt square: X² - 10x = 55 How to do that?

[Barbra]

X² - 10x = 55
How to do that?

 

[Barbra]

How can i let (X) alone

 

[BigBeachBanana]

How can i let (X) alone

It's easier to leave it in factor form and the square root of both sides.

[math]\begin{aligned}(x-5)^2 = 80\\ \sqrt{(x-5)^2} = \sqrt{80}\\ |x-5| = \sqrt{80} \end{aligned}[/math]

 

[Barbra]

It's easier to leave it in factor form and the square root of both sides.

[math]\begin{aligned}(x-5)^2 = 80\\ \sqrt{(x-5)^2} = \sqrt{80}\\ |x-5| = \sqrt{80} \end{aligned}[/math]

But i need it in Sum equation

 

[BigBeachBanana]

But i need it in Sum equation

Please start with the full and complete question next time. The ones you presented in posts #1 and #4 are 2 different questions.

[math]\sum_{i=1}^{10}(x_i-5)^2 =\sum_{i=1}^{10}x_i^2-10x_i+25= \sum_{i=1}^{10}x_i^2 - 10\sum_{i=1}^{10}x_i+ \sum_{i=1}^{10}25[/math]
Can you evaluate each of the second and third sum individually?

 

[Barbra]

I tried.. But i don't know how to continue
Are these steps correct?
And how to do it

 

[Barbra]

Ohhh yeah i got it
I will do it next times
Thank you

 

[Steven G]

X² - 10x = 55
How to do that?

First of all, you never asked a question. Possibly you want to know if x=11 is a solution, or maybe if x=4 is a solution or what is the value of x+2 or maybe x^2.

I suppose you want to solve for x in x^2-10x=55.
I would notice that x^2-10x factors as x(x-10).
Now x is some number while x-10 is 10 less than that number.
So the problem becomes to find two numbers where one number is 10 less than the other number and whose product is 55.

One set of solutions is sqrt(80)+5 and sqrt(80) - 5.
Can you find another set of numbers??

 

[Steven G]

After reading a later post from you I realize that your 1st post and later post are not the same question at all.