Find a number that has four identical factors each factor must be greater than 1 and can have only 1 and itself as factors
1 answer:
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We are given these three people age below:
- Jim's age
- Carla's age
- Tomy's age
We define the age of Jim as any variable, because the problem doesn't give any specific age. I will define Jim's age as x.
- Jim's age = x
Next, Carla is 5 years older than Jim. That means if Carla is 5 years older, her age would be x+5.
- Carla's age = x+5
Then Tomy is 6 years older than Carla. That means the age would be 6+(x+5).
- Tomy's age = 6+(x+5)
The sum of their three ages is 31 years old. That means we add all these ages and equal to 31.
Combine like terms and solve for x.
Then we substitute the value of x in ages to find these three people ages.
- Jim's age = x = 5
- Carla's age = x+5 = 5+5 = 10
- Tomy's age = 6+(x+5) = 6+(5+5) = 6+10 = 16.
Answer
- Jim's age = 5
- Carla's age = 10
- Tomy's age = 16
Answer:
There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.
Step-by-step explanation:
We notice that the expression on the left of the equation is a quadratic with leading term , which means that its graph is that of a parabola with branches going up.
Therefore, there can be three different situations:
1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.
2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.
3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.
So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.
We recall that the x-position of the vertex for a quadratic function of the form is given by the expression:
Since in our case and
, we get that the x-position of the vertex is:
Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:
This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.
We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:
See the graph produced in the attached image.
