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Algebra

Cool goIITian

Joined:	9 Oct 2007
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9 Oct 2007 11:00:25 IST

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HElp me with these 5 good questions..plz give the full solution

None

THe number of all possilbe integral values of x for which xsquare +19x+88 is a perfect square

If y= (sin^4 x -cos^4 x +sin^2 x cos^2x )/(sin^4 x+ cos^4 x +sin^2 x cos^2 x)
and x(0, 90degrees) then find the range of y

Find the real values of a for which equation x(x+1)(x+a+1)(x+a)=a^ 2(sqre) has 4real root

IF xsquare -x + a < 0 for atleast one negetive value of x then complete set values of a is

options......
1) (- ,3)
2) (- ,4)
3)(- ,2)
4) (- ,1)

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Comments (4)

Bipin Dubey

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Joined: 23 Jan 2007

Posts: 7942

9 Oct 2007 11:49:47 IST

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x² + 19x + 88 = (x+8)(x+11)

(x+11) - (x+8) = 3, so we have to find two numbers whose difference is 3 and when multiplied yeild a perfect square.

1. (-1) - (-4) = 3 and (-1)(-4) = 4 for which x = -12.
2. 4 - 1 = 3 and (1)(4) = 4 for which x = -7.

and for x = -8, -11 the expression reduces to 0 which is a perfect square.

Hence possible values of x are -12, -11, -8, -7.

Please post your other queries on a new page.

viswesh Nrayanan

Cool goIITian

Joined: 9 Oct 2007

Posts: 73

9 Oct 2007 17:01:36 IST

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Cool man...........HOw did it stike you...neyas .......lol.............I just found -8 and -11 and -12 but dint find the other -7 cool1!!!!!!!!!help me with other questions also

Gaurav |spideyunlimited| Ragtah

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Joined: 16 Dec 2006

Posts: 3373

10 Oct 2007 22:04:45 IST

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the method is as follows.
first factorize:
x^2 + 19x + 88
= (x+8)(x+11)

One solution is straightaway that either x = -8 or x = -11 as the product will become 0 which is a perfect square.

Second, for it to be a perfect square other than 0, the 2 numbers must have a diff. of 3 and their product must be a perfect square.
write numbers from 1 to 10 , and write the product formed by them and numbers 3 more than them.
1,4 = 4 ... 2,5 = 10 ....3,6 = 18....4,7 = 28.... upto 10,13 = 130
thus we see that the only case is 1,4 = 4
thus x+8 = 1 and x+11 = 4 gives x = -7
and x+8 = 4 and x+11 = 1 gives x = -12

Thus the solutions are x = {-7, -8, -11, -12}

*PLZ RATE*

Gaurav |spideyunlimited| Ragtah

Moderator

Joined: 16 Dec 2006

Posts: 3373

10 Oct 2007 22:13:56 IST

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IF xsquare -x + a < 0 for atleast one negetive value of x then complete set values of a is
ur options are wrong because
Discriminant of equation will be
1 - 4a.
which will be -ve for positive a
a can only have a max value of 1/4.
(and of course minimum value is - infinity)

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