Statement 1: (x² - 2x + A) > 0 for all x.prachich1987 wrote:Is A positive?
1) x^2-2x+A is positive for all x
2) Ax^+1 is positive for all x ... I assume it is Ax^2 + 1
Thanks Anurag.I understand that B alone is not sufficientAnurag@Gurome wrote:
Statement 1: (x² - 2x + A) > 0 for all x.
=> (x² - 2x + 1 + A - 1) > 0
=> (x - 1)² + (A - 1) > 0
=> (A - 1) > 0 ...................... As (x - 1)² ≥ 0 for all x
=> A > 1
=> A is positive
Sufficient
Statement 2 says (x² + 2x + A) > 0 for all x.prachich1987 wrote:Thanks Anurag.I understand that B alone is not sufficient
But according to me also A alone is not sufficient.
Refer to your explanation for statement 1 above
(x - 1)² + (A - 1) > 0
Assume that x=8
So the eqn would be 49+A>0
But it's not necessary that A >O
Assume that A=-7.The inequality is true.
Please advise if I am going wrong anywhere
What if x=0.5 & A= -0.5Anurag@Gurome wrote:Statement 2 says (x² + 2x + A) > 0 for all x.prachich1987 wrote:Thanks Anurag.I understand that B alone is not sufficient
But according to me also A alone is not sufficient.
Refer to your explanation for statement 1 above
(x - 1)² + (A - 1) > 0
Assume that x=8
So the eqn would be 49+A>0
But it's not necessary that A >O
Assume that A=-7.The inequality is true.
Please advise if I am going wrong anywhere
You've assumed A = -7, thus the expression becomes (x² + 2x + A) = (x² + 2x - 7). Now what for x = 0? (x² + 2x - 7) = -7, which is less than zero.
I think there has been a typo here, the equation is (x^2-2x+A) and not (x^2+2x+A).prachich1987 wrote:What if x=0.5 & A= -0.5Anurag@Gurome wrote:Statement 2 says (x² + 2x + A) > 0 for all x.prachich1987 wrote:Thanks Anurag.I understand that B alone is not sufficient
But according to me also A alone is not sufficient.
Refer to your explanation for statement 1 above
(x - 1)² + (A - 1) > 0
Assume that x=8
So the eqn would be 49+A>0
But it's not necessary that A >O
Assume that A=-7.The inequality is true.
Please advise if I am going wrong anywhere
You've assumed A = -7, thus the expression becomes (x² + 2x + A) = (x² + 2x - 7). Now what for x = 0? (x² + 2x - 7) = -7, which is less than zero.
x² + 2x + A= 0.25+1-0.5=0.75
Please try to understand the fact that in the expression (x² + 2x + A), x is a variable. You cannot determine the value of A for which (x² + 2x + A) > 0 for all x by fixing a value of x. That destroys the varying nature of x.prachich1987 wrote:What if x=0.5 & A= -0.5
x² + 2x + A= 0.25+1-0.5=0.75
Anurag@Gurome wrote:Statement 1: (x² - 2x + A) > 0 for all x.prachich1987 wrote:Is A positive?
1) x^2-2x+A is positive for all x
2) Ax^+1 is positive for all x ... I assume it is Ax^2 + 1
=> (x² - 2x + 1 + A - 1) > 0
=> (x - 1)² + (A - 1) > 0
=> (A - 1) > 0 ...................... As (x - 1)² ≥ 0 for all x
=> A > 1
=> A is positive
Sufficient
: : : As I have mentioned in the previous post (Just previous to you), for any negative value of A, the expression (x² + 2x + A) will be negative too for x = 0.goyalsau wrote:(x² - 2x + A) > 0 for all x
Lets Put X = -2 , A = -1
4 + 4 - 1 = 7
What's wrong in this???????:?:
Expression is still positive.......
Thanks Anurag,anshumishraAnurag@Gurome wrote:As I have mentioned in the previous post (Just previous to you), for any negative value of A, the expression (x² + 2x + A) will be negative too for x = 0.goyalsau wrote:(x² - 2x + A) > 0 for all x
Lets Put X = -2 , A = -1
4 + 4 - 1 = 7
What's wrong in this???????:?:
Expression is still positive.......
Put x = 0 with A = -1, the expression = -1 < 0
