[shivam_01]

Originally Posted by masoom

hey i am getting a different answer..

again discriminant should be a perfect square..

a^2- 64 = k^2..

(a-k) ( a+k) = 64..

RHS has 7 factors.. so factors will exhibit symmetry about the middle factor which is 8..

now a-k= 1 and a+k=64 will not give a integer..

a-k=2 and a+k=32 gives a=17..

a-k=4 and a+k=16.. gives a=10..

a-k=8 and a+k=8.. gives a=8..

so i think 3 values are possible..

Aisa kahi nahi likha hain ki -ve values not possible bhai
so a=8, -8 , 17 , - 17 , -10 ,10
i think

[Love_CAT]

8. If ax ^ 2 - bx + c = 0 has two distinct real roots in (0, 1), where a, b, c are natural numbers, then 16c(a-b+c)
(a) =a ^ 2 (b) < a ^ 2 (c) > a ^ 2 (d) ≥ a ^ 2

9. Number of possible value(s) of integer ‘a’ for which the quadratic equation x ^ 2+ ax + 16 = 0 has integral roots, is
(a)4 (b) 6 (c) 2 (d) none of these

10.I am able to solve this problem...Can anyone tell me what is the funda of sign change..

Thanks in advance

[masoom]

Originally Posted by Love_CAT

8. If ax ^ 2 - bx + c = 0 has two distinct real roots in (0, 1), where a, b, c are natural numbers, then 16c(a-b+c)
(a) =a ^ 2 (b) < a ^ 2 (c) > a ^ 2 (d) ≥ a ^ 2

9. Number of possible value(s) of integer ‘a’ for which the quadratic equation x ^ 2+ ax + 16 = 0 has integral roots, is
(a)4 (b) 6 (c) 2 (d) none of these

10.I am able to solve this problem...Can anyone tell me what is the funda of sign change..

Thanks in advance

the funda is that ...

if a polynomial..

f(x)= a1x^n + a2 x^n-1 +......+ an x+ an+1...

if we travel from the coefficient of the highest term..

to the constant...

we may encounter the chnage in signs of coeffcients..

the number of sign chnages gives the max possible +ve roots of f(x)..

if we repeat the same thing for f(-x)...

we obtain the number of -ve roots of f(x)...

hope its clear now..

[im_adi]

MORE QUESTIONS IN pdf or .doc format possible for u ???? pls upload here for benefit of all thanks !

[jain12345]

Originally Posted by im_adi

MORE QUESTIONS IN pdf or .doc format possible for u ???? pls upload here for benefit of all thanks !

yeah sure i will do that...