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Trignometry
9 Oct 2007 01:08:49 IST
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aNdRoMeDa
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Joined: 10 Sep 2007
Posts: 1319
9 Oct 2007 08:51:50 IST
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is d answer (-@,8) u (11,@)
@ = infinity
u= union
.............do tell if i m right.........
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9 Oct 2007 11:17:01 IST
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hey dude the question is dam good.
ok here is a solution------
x^2+19x+88=(x+11)(x+8)
now for it to be a square either x+11=x+8 or x+11 & x+8 both has to be square
case 1--- we get 11=8 absurd
case2 --- job at hand is simply to find 2 nos whith squares having difference of 3(as 11- 7 =3)
this possible only with (1,2) &(-1,-2)
hence required values of x = -7 &-12
ok here is a solution------
x^2+19x+88=(x+11)(x+8)
now for it to be a square either x+11=x+8 or x+11 & x+8 both has to be square
case 1--- we get 11=8 absurd
case2 --- job at hand is simply to find 2 nos whith squares having difference of 3(as 11- 7 =3)
this possible only with (1,2) &(-1,-2)
hence required values of x = -7 &-12
11
9 Oct 2007 12:09:32 IST
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ok here is a solution------
x^2+19x+88=(x+11)(x+8)
now for it to be a square either x+11=x+8 or x+11 & x+8 both has to be square
case 1--- we get 11=8 absurd
case2 --- job at hand is simply to find 2 nos whith squares having difference of 3(as 11- 7 =3)
this possible only with (1,2) &(-1,-2)
hence required values of x = -7 &-12
x^2+19x+88=(x+11)(x+8)
now for it to be a square either x+11=x+8 or x+11 & x+8 both has to be square
case 1--- we get 11=8 absurd
case2 --- job at hand is simply to find 2 nos whith squares having difference of 3(as 11- 7 =3)
this possible only with (1,2) &(-1,-2)
hence required values of x = -7 &-12
10 Oct 2007 21:59:02 IST
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the method is as follows.
first factorize:
x^2 + 19x + 88
= (x+8)(x+11)
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*
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