[jain12345]

Originally Posted by shivam_01

condition is b^2 >= 4ac Real roots

hence when b = 2 => a,c =1 1 value
=> b = 3 a,c = (1,1) (1,2) (2,1) ... so on

total 25 values

No its not 25

[jain12345]

guys here goes the next one

5. If a, b, c Є R and quadratic equation x ^ 2 + (a + b) x + c = 0 has no real roots then
(a) c (a + b + a) > 0

(b) c + c (a + b + c) > 0

(c) c + c (a + b – c)

(d) c (a + b – c) > 0

[the_hate]

Originally Posted by jain12345

guys here goes the next one

5. If a, b, c Є R and quadratic equation x ^ 2 + (a + b) x + c = 0 has no real roots then
(a) c (a + b + a) > 0

(b) c + c (a + b + c) > 0

(c) c + c (a + b – c)

(d) c (a + b – c) > 0

assuming a=0,b=0 and c=1,taking option c as >0

x^2+1 =0 has no real roots...
from options....option 2 is the only one that holds true..

[jain12345]

Originally Posted by the_hate

assuming a=0,b=0 and c=1,taking option c as >0

x^2+1 =0 has no real roots...
from options....option 2 is the only one that holds true..

you got it correct

[jain12345]

6. the equation (x^2+x+1)^2 + 1 =(x^2+x+1)(x^2-X-5) for x Є (-2,3) will have number of solutions.

(a) 1

(b) 2

(c) 3

(d) Zero

[jain12345]

6. The equation ax ^ 2 + bx + c = 0 and x ^ 2 + 2x + 3 = 0 have one root in common, then
(a) a : b : c = 1 : 2 : 3

(b) a : b : c = 3 : 2 : 1

(c) a : b : c = 1 : 3 : 2

(d) none of these

[shivam_01]

Originally Posted by jain12345

6. the equation (x^2+x+1)^2 + 1 =(x^2+x+1)(x^2-X-5) for x Є (-2,3) will have number of solutions.

(a) 1

(b) 2

(c) 3

(d) Zero

The answer is zero. No root in (-2,3)

[shivam_01]

The equation ax ^ 2 + bx + c = 0 and x ^ 2 + 2x + 3 = 0 have one root in common, then
(a) a : b : c = 1 : 2 : 3

(b) a : b : c = 3 : 2 : 1

(c) a : b : c = 1 : 3 : 2

(d) none of these


answer in bold

[Love_CAT]

The equation ax ^ 2 + bx + c = 0 and x ^ 2 + 2x + 3 = 0 have one root in common, then
(a) a : b : c = 1 : 2 : 3

(b) a : b : c = 3 : 2 : 1

(c) a : b : c = 1 : 3 : 2

(d) none of these

Answer is in bold

[Love_CAT]

6. the equation (x^2+x+1)^2 + 1 =(x^2+x+1)(x^2-X-5) for x Є (-2,3) will have number of solutions.

(a) 1

(b) 2

(c) 3

(d) Zero

Answer is in bold.