Find the integral values of x for which x2+19x+46 is a perfect square!
the soln is 35,-54,-25,6!!!
please try this qn-not easy mind you!!!!kvs
Write your expression as Undetermined error: @@(x+7)^2@@ + (5x-3). Put this equal to Undetermined error: @@(x+p)^2@@. You will get x= Undetermined error: @@(p^2-46)/(19-2p)@@ . It is clear from inspection that x will be an integer only if denominator of RHS is 1,-1,3,-3 ie. x=6,35,-25,-54. Hope you got it!
Nice method Vaibhav. But i don't get how it is clear from the inspection that x will be an integer only if denominator of RHS is 1,-1,3,-3 ie. Can you please explain this in some detail.
can a quadratic eqn have more than two roots? how?
ANS:x2+19x+46
=x2+18x+81+x-35
=(x+9)2+(x-35)
for the above equation to become perfect square x-35 must be equal to zero
therefore x=35
similarly, x2+19x+46=x2+20x+100-(x+54)
=(x+10)2-(x+54)
here x+54=0
=> x=-54
x2+19x+46=x2+22x+121-3(x+25)
=(x+11)2-3(x+25)
here x+25=0
=> x=-25
x2+19x+46=x2+16x+64+3x-18
=(x+8)2+3(x-6)
here x-6=0
=> x=6
Well sorry to say mate, no one is going to give u the solns in the exam for u to break the quadriatic in the manner u hav done........so i dont think this is the right method to do it!
hi amit- this qn is not a quadriatic eqn because for it to be a quadriatic eqn it must be equated to zero which is not the case here!!
PLZ CAN SOMEONE GIVE A MORE ACCEPTABLE ANSWER WHICH CAN B APPLIED TO SUCH QNS WHEN THE ANSWER IS NOT GIVEN!!!!
SORRY IF MY POST APPEARS RUDE TO ANYONE!
THANKS TO ALL FOR TRYING!KVS
Brilliant Charan . Great manupilation skills.
Note to kvs: Dude, this kind of question does involve a bit of manupilation to the best of my knowledge at this level. The method i propose has very little manupilation but nevertheless does have manupilation.
Let x2+19x+46=n2 where n is an integer.
Discriminant of the abv equation must be a perfect square and an integer for x to be integral.
Let, 192 -4(46- n2 ) = p2 where p is an integer.
Therefore (p-2n)(p+2n)=177=1*177=3*59.
Therefore p-2n = 1 & p+2n=177 . Solving we have n1 = 44.
Since p and n are integers therefore 177 must be divided into integral factors.
Also p-2n = 3 & p+2n = 59. Solving we have n2 = 14.
Substitute the values of n in the equation x2+19x+46=n2 to get the required values of x.
hey man u r perfectly righto!!! i don't think u manipulated at all. These prob. are of National Mathematics level probs. And they seem to be manipulating type but always have a well defined soln.
Grt. job man!!!
Grt. job man!!!
Sorry for being late in replying. I will tell you how it is clear from inspection that if x= Undetermined error: @@(p^2-46)/(19-2p)@@ then x will be an integer iff the denominator(D)=1,-1,3,-3.
x= Undetermined error: @@(p^2-49)/(19-2p)@@ + Undetermined error: @@3/(19-2p)@@
=(p+7)(p-7)/(19-2p)+ 3/(19-2p).
The numerator of the first term would be an integer for all integral values of p.
The second term will be an integer iff D=1,-1,3,-3. It can be seen that the corresponding values of p will also make the first term an integer.Hence x will also be an integer.
I had thought that this was quite obvious so had used the term "on inspection". Sorry for any misunderstanding!
x= Undetermined error: @@(p^2-49)/(19-2p)@@ + Undetermined error: @@3/(19-2p)@@
=(p+7)(p-7)/(19-2p)+ 3/(19-2p).
The numerator of the first term would be an integer for all integral values of p.
The second term will be an integer iff D=1,-1,3,-3. It can be seen that the corresponding values of p will also make the first term an integer.Hence x will also be an integer.
I had thought that this was quite obvious so had used the term "on inspection". Sorry for any misunderstanding!
Sorry for having irritated u! U must forgive human beings less intelligent thatn u. Still one small doubt remains: Why acc. to ur brand of reasoning it is not possible for
(p+7)(p-7)/(19-2p) = f1 & 3/(19-2p)= f2 where f1 & f2 be fractions and yet for f1+f2=integer(n)