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17 May 2020, 18:43
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Hi everyone,
I encountered this problem on TTP Linear and Quadratic Equations: Quiz 1 Medium, #12. I do want to credit one of their employees who helped me, I guess I just need a little more information on why:
Stem:
What is the value of x^2 + y^2
Issue:
My initial thought process is to assume that the stem is then: (x + y)^2 which then could be multiplied out to be: x^2 + 2xy + y^2. Apparently this isn't possible, and I need help understanding why, which then leads me to statement 1:
1: xy= -4 --> My thinking here is to plug in the -4 to the 2xy to get -8, and then to add the 8 to the other side of the equation, making it: x^2 + y^2 = 8 and making the statement efficient, but was told since the equation is not present, that this is wrong.
I am still in the earlier stages of studying, and thus am beating my brain into submission. Doing so with logic as opposed to 'just cause' will be much more beneficial.
Can anyone kindly flesh this out for me in greater detail? I have attached an image of the question as well.
I encountered this problem on TTP Linear and Quadratic Equations: Quiz 1 Medium, #12. I do want to credit one of their employees who helped me, I guess I just need a little more information on why:
Stem:
What is the value of x^2 + y^2
Issue:
My initial thought process is to assume that the stem is then: (x + y)^2 which then could be multiplied out to be: x^2 + 2xy + y^2. Apparently this isn't possible, and I need help understanding why, which then leads me to statement 1:
1: xy= -4 --> My thinking here is to plug in the -4 to the 2xy to get -8, and then to add the 8 to the other side of the equation, making it: x^2 + y^2 = 8 and making the statement efficient, but was told since the equation is not present, that this is wrong.
I am still in the earlier stages of studying, and thus am beating my brain into submission. Doing so with logic as opposed to 'just cause' will be much more beneficial.
Can anyone kindly flesh this out for me in greater detail? I have attached an image of the question as well.
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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18 May 2020, 00:03
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Please re-post the question according to our rules: rules-for-posting-please-read-this-before-posting-133935.html Thank you.
THIS TOPIC IS LOCKED AND ARCHIVED.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
