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olya-2409 [2.1K]
4 years ago
12

What are all the real zeroes of y = (x –12)^3 – 7

Mathematics
2 answers:
kodGreya [7K]4 years ago
8 0

y = (x –12)^3 – 7

set y=0

0 = (x –12)^3 – 7

add 7 to each side

7  = (x –12)^3

take the cube root of each side

7^ 1/3 = x-12

add 12 to each side

12 + 7^(1/3) = x

this is the only real root

the other 2 are imaginary

Delvig [45]4 years ago
3 0

You might be right.

Let y = 0

(x - 12)^3 - 7 = 0         Subtract 7 from both sides

(x - 12)^3 = 7              Take the cube root of both sides.

x - 12 = cuberoot(7)   Add 12 to both sides

x = 12 + cuberoot(7)

Are the other two real or imaginary? The quickest way to find out the answer is to graph the original equation. It has the shape of something that crosses the x axis but once. So the other two roots are imaginary.

My calculator says that the real root is 13.91

The two complex ones are 11.04 +/- 1.66i which of course is not real.

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