48. How many real numbers satisfy x^6+|x|=7?

that is always + because even power is + so if we find a plus x, its negative mate will also work

x^6 + x = 7
x(x^5 + 1)= 7
if 0<x<1 no way
if x = 2, way too big
so between 1 and 2
1.2 ---> 4.18
1.3 ---> 6.12
So there is one between 1 and 2 that works, and its negative mate
so two

X > 10 how many numbers satisfy

To find the number of real solutions to the equation x^6 + |x| = 7, we can analyze different cases.

Case 1: x is positive (x > 0)
Since |x| is always positive, we can rewrite the equation as x^6 + x = 7. However, it is difficult to solve this equation algebraically. We can use numerical methods to find possible solutions.

Case 2: x is negative (x < 0)
When x is negative, the absolute value |x| becomes positive, so we can rewrite the equation as x^6 - x = 7. As in case 1, this equation is difficult to solve algebraically, but we can use numerical methods to find possible solutions.

Since cases 1 and 2 both require numerical methods to find solutions, we will not be able to determine the exact number of real solutions without using these methods. However, it is possible to analyze the behavior of the equation graphically.

By graphing the equation y = x^6 + |x| - 7, we can observe the number of intersections with the x-axis, which represents the number of real solutions.

Therefore, to determine the exact number of real solutions, we would need to use graphing software or a graphing calculator.

To determine the number of real solutions for the equation x^6 + |x| = 7, we need to carefully analyze the equation.

First, let's examine the term |x|. The absolute value function |x| disregards the negative sign of x and returns a positive value. Hence, |x| is always non-negative, equal to x when x is positive, and equal to -x when x is negative.

Now, let's consider the possible values for x^6. Since any real number raised to an even power is non-negative, x^6 is always non-negative regardless of the value of x. Hence, the left-hand side of the equation (x^6 + |x|) is non-negative as well.

Given that the equation is x^6 + |x| = 7, we have a condition: x^6 + |x| must be equal to a positive number (7 in this case). Since we know that x^6 + |x| is non-negative, it cannot be negative. Therefore, there cannot be any negative real solutions for this equation.

Now, let's consider the possible scenarios for x^6:

1. If x^6 = 0, then |x| (which is either 0 or -x) must be equal to 7. However, this is not possible since 0 and -x (where x is not 0) are less than or equal to 0. Therefore, x^6 = 0 does not satisfy the equation.

2. If 0 < x^6 < 7, then |x| (which is x) must be equal to 7 - x^6. However, 0 < x^6 < 7 implies 0 < 7 - x^6 < 7. But the absolute value of a positive number is always positive, so |x| = x cannot be equal to 7 - x^6, a positive number. Therefore, x^6 in this range does not satisfy the equation.

3. If x^6 = 7, then |x| (which is either x or -x) must be equal to 0. However, this contradicts the fact that 0 is not equal to 7. Therefore, x^6 = 7 does not satisfy the equation.

4. If x^6 > 7, then |x| (which is x) must be equal to x^6 - 7. Since x^6 > 7 implies x^6 - 7 > 0, |x| = x is greater than 0. Thus, we have a possible solution here.

In conclusion, there is exactly one real number solution for the equation x^6 + |x| = 7.