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imslogic
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Square Root Inequalities 08 Dec 2021, 00:04
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Question: (x-1)^(1/2) + 3 > x

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Square Root Inequalities 08 Dec 2021, 03:14
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Question: (x-1)^(1/2) + 3 > x

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VeritasKarishma

Dear Karishma: Can you please advise using the wavy line method? I get an answer of 2<x<5 but was told that the answer is 1<=x<5. Thanks!
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Square Root Inequalities 09 Dec 2021, 00:05
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imslogic
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Question: (x-1)^(1/2) + 3 > x

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VeritasKarishma

Dear Karishma: Can you please advise using the wavy line method? I get an answer of 2<x<5 but was told that the answer is 1<=x<5. Thanks!
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Hey imslogic! This post is likely to get locked because it does not follow the rules of posting in the forum.
Please put up the complete question with the options and relevant tags in the Problem Solving forum.
Btw, the answer is 2 < x < 5. Note that the inequality holds for neither 1 nor 2.
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Square Root Inequalities 10 Dec 2021, 05:26
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imslogic
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Question: (x-1)^(1/2) + 3 > x

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VeritasKarishma

Dear Karishma: Can you please advise using the wavy line method? I get an answer of 2<x<5 but was told that the answer is 1<=x<5. Thanks!

Hey imslogic! This post is likely to get locked because it does not follow the rules of posting in the forum.
Please put up the complete question with the options and relevant tags in the Problem Solving forum.
Btw, the answer is 2 < x < 5. Note that the inequality holds for neither 1 nor 2.
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Hi Karishma: Thank you for your response. I will try to find the full question but that level of information is all I have thus far. You are the best tutor on GMATCLUB.
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Square Root Inequalities 20 Dec 2021, 04:12
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VeritasKarishma

Dear Dr. Karishma: I have another question for you. How does one solve a^3 + b^3? I understand that the answer is (a+b)^3 - 3ab(a+b), however I was earlier taught that x^3 + y^3 = (x+y)(x^2+y^2-xy). Please clarify. Thank you!

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Square Root Inequalities 20 Dec 2021, 21:00
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VeritasKarishma

Dear Dr. Karishma: I have another question for you. How does one solve a^3 + b^3? I understand that the answer is (a+b)^3 - 3ab(a+b), however I was earlier taught that x^3 + y^3 = (x+y)(x^2+y^2-xy). Please clarify. Thank you!

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This is the algebraic identity:

\((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)

This leads to
\(a^3 + b^3 = (a + b)^3 - 3ab(a + b)\)


If you feel confused about it, you can simply open the brackets yourself to see what you get:
\((a + b)^3 = (a+b)(a+b)(a+b) \)
\(= (a^2 + 2ab + b^2)(a + b) \)
\(= a^3 + a^2b + 2a^2b + 2ab^2 + b^2a + b^3 \)
\(= a^3 + 3a^2b + 3ab^2 + b^3 \)
\(= a^3 + b^3 + 3ab(a + b)\)
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Square Root Inequalities 22 Dec 2021, 04:25
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KarishmaB



This is the algebraic identity:

\((a + b)^3 = a^3 + b^3 + 3ab(a + b)\)

This leads to
\(a^3 + b^3 = (a + b)^3 - 3ab(a + b)\)


If you feel confused about it, you can simply open the brackets yourself to see what you get:
\((a + b)^3 = (a+b)(a+b)(a+b) \)
\(= (a^2 + 2ab + b^2)(a + b) \)
\(= a^3 + a^2b + 2a^2b + 2ab^2 + b^2a + b^3 \)
\(= a^3 + 3a^2b + 3ab^2 + b^3 \)
\(= a^3 + b^3 + 3ab(a + b)\)
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Thank you, Karishma. Your response was most helpful. So, both equations stated in my query - a^3 + b^3=(a+b)^3 - 3ab(a+b) and x^3 + y^3 = (x+y)(x^2+y^2-xy) - are the same equations stated differently. The first is in expanded form whereas the second is in factored form. Thank you!

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Hi there,
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