<p>sinX + cosX= 1,4
what is sinXcosX ???</p>
<p>if 4+radical2 is one root of a quadratic equation given by x^2-Px+Q=0 where P and Q are rational numers then P is...???</p>
<p>And finally what are ARBITRARY INTEGERS????</p>
<p>sinX + cosX= 1.4
(sinX + cosX)² = sin²X + 2sinXcosX + cos²X = 1.4² = 1.96
2sinXcosX + 1 = 1.96
sinXcosX = 0.48</p>
<p>4-sqrt(2) since roots involving radicals must occur in conjugates.</p>
<p>Thanks for the fast response..</p>
<p>Do you for certain the first answer is correct???
This question was a typo of Meylani's book, which he corrected later and put the right answer on the web. On internet the correct answer is 0,96...
So if we are correct, then he made a mistake with correcting the typo</p>
<p>Btw.. what is the answer for the question involving radicals??</p>
<p>the equation you wrote has two root 4 + rad2 and another one call it y.
Now 4+rad2 + y=P by Vieta's formulas where P sis rational so we expect
4+rad2 + y to be rational too that meaning y must be of the form A - rad2 where A is ratonal.
Also we know that (4+rad2)(A-rad2)=Q expanding we get
4A -4rad2 +Arad2 -2=Q so A must be 4 for the expression in the left to be rational.
This is the long way but you may want to memorize that if one root of a quadratic equation is of the form a+radicalB the other will be a-radicalB and vice versa.</p>
<p>Yes, I'm sure that the answer is 0.48. If you solve sinX + cosX = 1.4, you get x = 0.927295218. Then sinXcosX = 0.48.</p>
<p>Ok thank you for the explanation...
But what is the answer of the radical question, you need the find P.</p>
<p>And any ideas about arbitrary integer??</p>
<p>Deference:
TemplarOfSteel gave you an excellent explanation, but you seem to be unable to recognize <em>any</em> answer except 'P= ...' . </p>
<p>Try this, from Templar's explanation:
P = 4 + sqrt(2) + y
where y = A - sqrt(2)
and A = 4</p>
<p>Solve first for y, and then for P. It's simple. Don't look to be spoonfed.</p>
<p>SOrry if I I look to be spooned</p>
<p>The right answer is 8, but you all give 4 as answer</p>
<p>Deference:
Sigh....</p>
<p>All right, here's the answer:
Substitute A=4 into y = A - sqrt(2) to get y = 4 - sqrt(2)
Substitute this value of y into
P = 4 + sqrt(2) + y to get
P = 4 + sqrt(2) + 4 - sqrt(2)
or P = 8</p>
<p>ok I understand now,
But do you really need to know this...
the 2 books that I use barrons and sparknotes never talked about this</p>
<p>Deference:
Sorry - I didn't mean to snarl at you :) . You caught me in a bad mood.</p>
<p>But seriously - you need more practice on problems like this, which do not fall into nice shrink-wrapped scenarios of use<em>this</em>formula.</p>
<p>Good luck on your SAT - keep on plugging' .</p>