My Math Forum Doubts !

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September 7th, 2012, 11:13 AM   #1
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Doubts !

Please clarify the following doubts of mine .

Q1 ) [attachment=1:142g6gcc]doubt.JPG[/attachment:142g6gcc]

Q2 )[attachment=0:142g6gcc]dout.JPG[/attachment:142g6gcc]

Thanks .
Attached Images
 doubt.JPG (11.6 KB, 89 views) dout.JPG (16.7 KB, 89 views)

 September 7th, 2012, 12:54 PM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Doubts ! Q1: $(a+b+c)^2= a^2+b^2+c^2+2(ab+ac+bc) = 7^2 = 49$ $a^2+b^2+c^2= 49 - 2(ab+ac+bc) = 49 - 40 = 9$ Q2: Ans c)
September 7th, 2012, 07:46 PM   #3
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Re: Doubts !

Quote:
 Originally Posted by saravananr Please clarify the following doubts of mine .
Those are not doubts...they are requests to do YOUR homework

 September 7th, 2012, 08:00 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Doubts ! One way to figure out the second problem is to write: $a^2+b^2+c^2-ab-bc-ca=$$\frac{1}{2}a^2-ab+\frac{1}{2}b^2$$+$$\frac{1}{2}b^2-bc+\frac{1}{2}c^2$$+$$\frac{1}{2}c^2-ca+\frac{1}{2}a^2$$=$ $\frac{1}{2}$$\(a^2-2ab+b^2$$+$$b^2-2bc+c^2$$+$$c^2-2ca+a^2$$\)=\frac{1}{2}$$(a-b)^2+(b-c)^2+(c-a)^2$$$
 September 7th, 2012, 08:15 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Doubts ! Another (similar) approach to the first problem: $a+b=7-c$ (1) $a^2+2ab+b^2=7^2-14c+c^2$ $a+c=7-b$ (2) $a^2+2ac+c^2=49-14b+b^2$ $b+c=7-a$ (3) $b^2+2bc+c^2=49-14a+a^2$ Adding (1), (2) and (3) we obtain: $2a^2+2b^2+2c^2+2(ab+ac+bc)=147-14(a+b+c)+a^2+b^2+c^2$ $a^2+b^2+c^2=147-14\cdot7-2\cdot20=9$
 September 7th, 2012, 08:18 PM #6 Member   Joined: Sep 2012 Posts: 32 Thanks: 0 Re: Doubts ! Thank you very much for your kind and quick reply. You have cleared all my doubts.

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