Bunuel wrote:
unraveled wrote:
Squaring the given expression, weobtain:
- \( \sqrt{c - x} = \sqrt{c} -\sqrt{x}\)
\( c - x =c - 2\sqrt{cx} +x\)
\( x =\sqrt{cx}\)
Now, squareagain:
- \( x^2 =cx\)
\( x^2 - cx =0\)
\( x(x - c) =0\)
The above equation has two solutions, x = 0 and x =c.
Answer:C.
Hi Bunuel,
As x = c so any nonnegative value of c is value of x. Thu,
Should not the answer be Ethen?
Thanks.
c there is a fixed constant, not a variable. So it has one fixed value, say 10 or12.[/quote]
yes, i realised it later after a couple of minutes of posting the question. Silly me.
Only that if someone tries testing numbers on hit and trial basis, he/she is going to get wrong answer.
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Statistics : Posted by unraveled • on 17 Feb 2024, 06:59 • Replies 6 • Views 233
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