How do you solve and write the following in interval notation: #x^2-1x-30>0#?

Answer 1

Solution : # x <-5 or x >6 #. In interval notation: #(-oo , -5) uu (6,oo)#.

#x^2-x-30 >0 or x^2-6x+5x-30 >0 or x(x-6) +5(x-6) >0 or (x+5)(x-6)>0# Critical points are #x = -5 , x =6 #
When # x < -5; (x+5)(x-6) is (-)*(-) = (+) or >0 #
When # -5< x < 6; (x+5)(x-6) is (+)*(-) = (-) or <0 #
When # x > 6; (x+5)(x-6) is (+)*(+) = (+) or >0 #
Solution : # x <-5 or x >6 #. In interval notation: #(-oo , -5) uu (6,oo)#. The graph also confirms the findings. graph{x^2-x-30 [-80, 80, -40, 40]}[Ans]
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Answer 2

To solve the inequality ( x^2 - x - 30 > 0 ), you first factor the quadratic expression ( x^2 - x - 30 ) to find its roots. The roots are ( x = -5 ) and ( x = 6 ). Then, you determine the intervals where the expression is greater than zero by testing intervals around the roots. The solution is ( x \in (-\infty, -5) \cup (6, \infty) ) in interval notation.

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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.

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.

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.

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.

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