solving inequalities?

Question

x^2-5x>0

I have attempted this problem FIFTEEN times, if anyone can help please show work.

has to be in interval notation

I'm not solving for x - the inequality is supposed to have that zero on one side

none of these are correct

Answer 1

x^2 - 5x > 0 means x(x - 5) > 0.

Thus, we obtain two inequalities:
(a) x > 0 and
(b) x - 5 > 0.

Observe that from these two inequalities, we have two critical numbers that will make the left-hand side of (a) and (b) to be either 0 or undefined. These are x = 0 for (a); and x = 5 for (b).

From these critical numbers, we split the real line into the following open intervals:
(–∞, 0), (0, 5) and (5, ∞).

Next, we check the behavior (in terms of signs) of the left-hand side of the inequalities (including the given inequality) together with the open intervals and critical numbers.

Intervals       x       x – 5       x(x – 5)
(–∞, 0)       –             –             +
x = 0       0             –            0
(0, 5)       +            –             –
x = 5       +             0             0
(5, ∞)       +            +             +

NOTE: Consider the open interval (0, 5). If we pick x on this interval, say x = 1. Then obviously on the x column, x = 1, so the sign of x is +, since 1 > 0. But on the x - 5 column, we have 1 - 5 = –4. So the sign on this column is – since –4 < 0. Thus, we take the product of these signs, i.e. (+)(–) = – that we will mark on the x(x – 5) column.

Notice that on the rightmost column ( for x(x – 5) ), we took the product of the signs located on the second and third column that are produced by the corresponding interval in the leftmost column. Since we want x(x – 5) > 0; i.e., the sign for x(x – 5) will yield +, we observe from the table that any value of x in the open intervals (–∞, 0) or (5, ∞) yields a + sign in the x(x – 5) column. Hence, the solution for this inequality must be the set
{x | x is in the interval (–∞, 0) or (5, ∞)}.

Answer 2

1. Find the roots of the equation y = x^2 -5x
Let y = 0
x(x-5) = 0
x = 5 and 0

2. Plot 5 and 0 on a number line

3. If x>5, is the inequality positive (>0)?
If 0 if x<0?

Pick a point greater than 5, say 6. Plug it back into the inequality (6^2 -5*6=6) which is greater than 0.

Pick a point between 0 and 5, say 1. Again, plug it in (1^2-5*1=-4) which is less than 0

Pick a point less than 0, say -1. (-1^2 -5*-1=6) which is greater than 0.

Thus, because points >5 and <0 made the inequality postive (>0, as seen in the inequality), the solution set is:

x>5 OR x<0 (it cannot be both at the same time)

Now we can change these two conditions to interval notation:
x>5 --> (5, +infinity)
x<0 --> (- infinity, 0)

Answer 3

x^2-5x>=0 => x^2 >= 5x ; from this x = 0 is a solution
divide by x
x>=5.

[5, INF) v {0} is the solution i n inteveral notation

Answer 4

x^2-5x > 0

Taking x common

x(x-5) > 0

for this inequlity to hold

x>0 or x>5

therefore one solution is x>5

otherwise

x<0 and x<5 ie x<5 will also hold for this

two possible solutions are x>0 and x<0

Answer 5

x^2-5x>0
x^2-5x+(25/4)>0+(25/4)
(x-(5/2))^2 > (25/4)
x - (5/2) > sqrt (25/4)
x - (5/2) > +/- (5/2)
x > [5/2 + 5/2] , x < [-5/2 + 5/2]
x > 5, x< 0
So plug in 6 and -1.

Answer 6

x^2-5x>0
add 5x to both sides
x^2-5x + 5x> 0+5x
x^2>5x
X^2 > 5x
divide both sides by x
Answer
X>5

Answer 7

add 5x to both sides: x^2 > 5x
divide both sides by x: x>5

Answer 8

Try this:

Divide each side by x, which will leave you with
x-5>0
Add 5 to both sides and solve to get x>5