Discussion :: Problems on Numbers - General Questions (Q.No.12)
|Kumar said: (Jul 17, 2011)|
|The question makes no sense yaar. What does the question maker mean by saying 'sum of their sqaures'. A sense English means x^2+y^2 , but the solution shows (x+y)^2 [20+3] when the given question should have been like 20^2+3^2. The question should have been 'square of their sum' and not 'sum of their squares'.|
|Abhishek Patel said: (Jul 19, 2014)|
|Can anybody make me understand, why in the given solution we are adding 240 in 289 whereas actually it should be deducted from 289.|
|Pavan said: (Jul 24, 2014)|
In this question we already know their products and sum of the squares of their products only one we have to find out is their sum. So here we are using (a+b)whole square formulae:
Hence we are adding 289+2(120).
|Mujahed said: (Dec 30, 2014)|
|Here ab = 120, a^2+b^2 = 289.
Very simple (a+b)^2 = a^2+b^2+2ab.
Here given a^2+b^2 = 289 & ab = 120.
= 289 + 2(120).
(a+b)^2 = 289 + 240.
(a+b)^2 = 529.
(a+b) = 23.
|Jayesh said: (May 22, 2016)|
|The two numbers are 15 and 8.
15 + 8 = 23.
15* 8 = 120.
225 + 64 = 289.
|Mohita said: (Jun 23, 2016)|
|Shouldn't the solution be -
x^2 + y^2 + 2xy = 289 + 2xy
If we are opening the (x+y)^2 bracket on one side of the equation, a 2xy is being added to the LHS which must also be compensated for on the RHS of the equation. Please correct me if I'm wrong.
|Ram said: (Jul 12, 2016)|
|x^2 + y^2 = 289 and we say it as x^2 + y^2 = 17^2 and take root at both sides.
It will give answer as 17. If this is correct or 23 is correct?
Please tell me.
|Pooja P Kumar said: (Apr 18, 2017)|
|You are absolutely wrong @Kumar.
The question is correct and the method used is perfect.
Just apply the formula (a+b)^2 = a^2+b^2+2ab, you will get the solution.
|Tirth said: (Sep 28, 2018)|
|120*2 = 240.
240+289 = 529.
√529 = 23.
|Sourav Yadav said: (Aug 19, 2019)|
|It clearly says that "sum of squares ". It does not say" square of the sum".
So basically considering this case the question should be carried out by doing x^2 + y^2 and not (x+y) ^2.
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