How do you solve # x^2-5x>0 #?

Answer 1
#x^2-5x>0# could be written as #x^2> 5x#
Case 1: x<0 then #x^2>0# and #5x<0# So the inequality is true for all values of #x<0#
**Case 2: x>=0#** Dividing both sides by #x# gives #x>5#
Complete Solution #x<0# or #x>5#
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Answer 2
To solve \( x^2 - 5x > 0 \), you first factor the expression to find the critical points. The critical points are where the expression equals zero, so factor \( x \) out: \( x(x - 5) > 0 \). Then, you solve for the critical points: \( x = 0 \) and \( x = 5 \). Now, you can test the intervals created by these critical points. Choose a test point from each interval and evaluate the inequality. For example, test \( x = -1 \) in the interval \( x < 0 \), which gives \( (-1)(-1 - 5) > 0 \), which is true. Then test \( x = 3 \) in the interval \( 0 < x < 5 \), which gives \( (3)(3 - 5) > 0 \), which is false. Finally, test \( x = 6 \) in the interval \( x > 5 \), which gives \( (6)(6 - 5) > 0 \), which is true. Thus, the solution to \( x^2 - 5x > 0 \) is \( x < 0 \) or \( x > 5 \).
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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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