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.
Answer:
A line has length, but no width or height
Step-by-step explanation:
I dont believe there are any
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Do 42 divided by 15 then divided 12 by the awnser. And you do 42 divided by 28 and divided that by 12 I think I'm pretty sure
