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Jadedry
 one year ago
Hi!
I'm having trouble with the following question:
"Solve the following quadratic formula by completing the square:
x^2 2x +a = 0 "
So far, I'm managed to find out that
a)The equation has 2 distinct roots because
"b^2 4ac" = positive
b) One of the roots is 1.
My textbook says that the other answer is sqrt(1a) but I can't get that answer. My calculations show that the answer should be (sqrt(1a) + 1 )
Could you please show working? Thanks in advance!
Jadedry
 one year ago
Hi! I'm having trouble with the following question: "Solve the following quadratic formula by completing the square: x^2 2x +a = 0 " So far, I'm managed to find out that a)The equation has 2 distinct roots because "b^2 4ac" = positive b) One of the roots is 1. My textbook says that the other answer is sqrt(1a) but I can't get that answer. My calculations show that the answer should be (sqrt(1a) + 1 ) Could you please show working? Thanks in advance!

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hartnn
 one year ago
Best ResponseYou've already chosen the best response.2so you know how to solve quadratics using completing the square method... can you just post first few steps of your working??

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2i believe you reached till \(\large x^2 2x+1 = 1a\) right ??

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2oh and just so you know, your answer of \(\large x = \sqrt{1a}+ 1 \) is correct. and the book is wrong

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2another answer is \(\large x = \sqrt{1a} +1 \)

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1I did indeed reach \[x^{2}2x +1 = 1a\] at which point it seemed that \[x1 = \sqrt{1a}\]  Glad to know that my answer was right though! I was ripping my hair out. Thanks for clarifying. c:

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2welcome ^_^ i see you're new here, \(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Jadedry
 one year ago
Best ResponseYou've already chosen the best response.1Haha, thank you for warm welcome! I'll probably be lurking around in these forums quite often. c:
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