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How do you solve #x^2+x+1>0 #?

Answer 1

#x in RR#

Consider the related equation #x^2+x+1=0#

Evaluating the discriminant (#Delta=b^2-4ac#) #color(white)("XXX")Delta = -3# and we know that if #Delta < 0# there are no Real solutions.

That is #x^2+x+1# does not touch or cross the X-axis.

We know that for some values (e.g. #x=0#) #color(white)("XXX")x^2+x+1 > 0# and since it doesn't touch or cross the X-axis for all Real values of #x# #color(white)("XXX")x^2+x+1 > 0#

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Answer 2

Hazel Anglin

To solve the inequality x^2 + x + 1 > 0, you can use various methods such as factoring, completing the square, or the quadratic formula. However, in this case, since the quadratic expression does not factor nicely and the discriminant is less than zero, indicating that the quadratic equation does not have real roots, we can conclude that the inequality x^2 + x + 1 > 0 holds true for all real values of x.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #x^2+x+1&gt;0 #?

How do you solve #x^2+x+1>0 #?