sch wrote:I have tough time with inequalities, so you can expect more posts from me. Right now im having problems solving these two:
2x^2+x</=1
x^2+x+1>0
can anyone explain step by step these two?

Can you edit your first equation. It is not clear (2x^2+x</=1)
For the 2nd equation :

x^2+x+1>0
=> (x+1/2)^2 + 3/4 > 0

Since the part in bold is always non-negative,hence the inequality will be always vallid, so x can take any value.

anshumishra wrote:

sch wrote:I have tough time with inequalities, so you can expect more posts from me. Right now im having problems solving these two:
2x^2+x</=1
x^2+x+1>0
can anyone explain step by step these two?

Can you edit your first equation. It is not clear (2x^2+x</=1)
For the 2nd equation :

x^2+x+1>0
=> (x+1/2)^2 + 3/4 > 0

Since the part in bold is always non-negative,hence the inequality will be always vallid, so x can take any value.

Hey Anshu is there any specific trick to convert x^2+x+1 to (x+1/2)^2 or it just by hit & trial.

ankur.agrawal wrote:

anshumishra wrote:

sch wrote:I have tough time with inequalities, so you can expect more posts from me. Right now im having problems solving these two:
2x^2+x</=1
x^2+x+1>0
can anyone explain step by step these two?

Can you edit your first equation. It is not clear (2x^2+x</=1)
For the 2nd equation :

x^2+x+1>0
=> (x+1/2)^2 + 3/4 > 0

Since the part in bold is always non-negative,hence the inequality will be always vallid, so x can take any value.

Hey Anshu is there any specific trick to convert x^2+x+1 to (x+1/2)^2 or it just by hit & trial.

It is not hit and trial:
Since, a^2 + 2ab + b^2 = (a+b)^2
So my intention is to extract a square out of x^2 + x + 1 in the same format :
here : x = a
2ab = x = 2*x*(think what to write here to make the multiplication equal to x) = 2*x*1/2
x^2+x+1 = [x^2 + 2*x*(1/2) + (1/2)^2] + 3/4 [Since (1/2)^2 = 1/4 and we had 1 in the equation, we need to add this 3/4]
=[x+(1/2)]^2 + 3/4

Hope that helps.

anshumishra wrote:

ankur.agrawal wrote:

anshumishra wrote:

sch wrote:I have tough time with inequalities, so you can expect more posts from me. Right now im having problems solving these two:
2x^2+x</=1
x^2+x+1>0
can anyone explain step by step these two?

Can you edit your first equation. It is not clear (2x^2+x</=1)
For the 2nd equation :

x^2+x+1>0
=> (x+1/2)^2 + 3/4 > 0

Since the part in bold is always non-negative,hence the inequality will be always vallid, so x can take any value.

Hey Anshu is there any specific trick to convert x^2+x+1 to (x+1/2)^2 or it just by hit & trial.

It is not hit and trial:
Since, a^2 + 2ab + b^2 = (a+b)^2
So my intention is to extract a square out of x^2 + x + 1 in the same format :
here : x = a
2ab = x = 2*x*(think what to write here to make the multiplication equal to x) = 2*x*1/2
x^2+x+1 = [x^2 + 2*x*(1/2) + (1/2)^2] + 3/4 [Since (1/2)^2 = 1/4 and we had 1 in the equation, we need to add this 3/4]
=[x+(1/2)]^2 + 3/4

Hope that helps.

U rock man.Thanks Again

Thank you for explanation. The first formula is correct </= means less or equal too. I guess the steps to derive the solution would be the same as if wer re-wrote theequation as 2x^2+x<1

sch wrote:Thank you for explanation. The first formula is correct </= means less or equal too. I guess the steps to derive the solution would be the same as if wer re-wrote theequation as 2x^2+x<1

great! We normally use <= for less than equal sign.
Let me know if you need help with this one.
Thanks!

sch wrote:I have tough time with inequalities, so you can expect more posts from me. Right now im having problems solving these two:
2x^2+x</=1
x^2+x+1>0
can anyone explain step by step these two?

Solution to 2nd equation :

2x^2+x<=1
=> 2x^2 + x - 1 <= 0
=> 2x^2 + 2x -x - 1<=0
=> (2x-1)(x+1) <= 0
=> -1 <= x <= 1/2