Question

the no. of integral values of a for which x^2-(a-1)x+3=0 has both roots positive and x^2+3x+6-a=0 has both roots negative is
a.0
b.1
c.2

Answer 1

consider the roots as 'a' and 'b' then in first part a>0 ,b>0 so a+b>0 => a-1>0 a>1
put d>=0
think of same codn in 2nd part

Answer 2

consider roots be r and t rather than a and b

Answer 3

kya kehna h?

Answer 4

mtlabb..

Answer 5

matlab samjhi ya nahi?

Answer 6

one equation is a^2-2a-13>=0

Answer 7

take intersection of both the inequalities

Answer 8

second equation is 4a-15>=0

Answer 9

Let p and q be the roots of x^2-(a-1)x+3=0.
For both roots to be positive (p+q) = (a-1) > 0; ==> a > 1 and pq = 3 > 0 which we already know it is.
But there is one more condition to be met. The discriminant must be positive too. There are cases where p + q > 0 and pq > 0 and yet the roots might be complex.
Discriminant = (a-1)^2 - 12 > 0 too. Find the intersection of those 2 inequalities for this equation.

Answer 10

how am i going to find intersection a>1 and a^2-2a-13>=0

Answer 11

(a-1)^2 - 12 > 0
(a-1)^2 > 12
(a-1) > sqrt(12)
a > sqrt(12) + 1

Answer 12

intersection wud be a>sqrt 12+1

Answer 13

yes.

Answer 14

do the similar procedure for x^2+3x+6-a=0 has both roots negative

Answer 15

then wt abt second one

Answer 16

i got value of a as a belongs to (15/4,6)

Answer 17

@ranga you'll get some other roots if you'll take sqrt by that way

Answer 18

x^2+3x+6-a=0
Assume r and s are the roots. r + s = -3 and rs = 6 - a
If both roots are negative r + s < 0 which it is because r + s = -3
rs should be positive. So (6-a) > 0 or a < 6.
Also the discriminant = 9 - 4(6-a) > 0.
Find the intersection of these two inequalities as well and then intersect with the previous solution.

Answer 19

i m also geting dis part only

Answer 20

@divu.mkr
(a-1)^2 > 12 ==> (a-1) > sqrt(12) or (a-1) < - sqrt(12). The second one can be thrown away because we already have one inequality where a > 1 and it will not intersect with (a-1) < - sqrt(12)

Answer 21

ohhhkk

Answer 22

1 + sqrt(12) = 4.46.
So 4.46 < a < 6
The only integer value for a is 5.

Answer 23

^^^ is this for a different problem or for the original posting?

Answer 24

same

Answer 25

That looks like a solution for x^2 - 2x - 11 = 0

Answer 26

yep...she was having a prob in solving that

Answer 27

thanku...for telling me the solution and bearing my madness for such a long tym.....
truly thanks....dhanyavaad

Answer 28

I had to look up dhanyavaad :)