Treat x^4 as the square of p: x^4 = p^2.
Then x^4 - 5x^2 - 36 = 0 becomes p^2 - 5p - 36 = 0.
This factors nicely, to (p-9)(p+4) = 0. Then p = 9 and p = -4.
Equating 9 and x^2, we find that x=3 or x=-3.
Equating -4 and x^2, we see that there's no real solution.
Show that both x=3 and x=-3 are real roots of x^4 - 5x^2 - 36 = 0.
For Letter C You Need to Count The Squares in the Pond And the Half Squares are a Half. The Answer for C is 33 Because The Half and the Other Half is a Whole So I Made 3 New Wholes and Added it to the 30.
Answer:
x=4
Step-by-step explanation:
Problem set up is: 5x-6=14
5x=20
x=4
Hello!!
Equation forms:

Therefore,

And, this is how your graph would look
Good luck :)