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

User Gohawks
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2 Answers

4 votes

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.

What are all the real zeroes of y = (x –12)^3 – 7-example-1
User Glena
by
8.2k points
5 votes

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

User Markus Rudel
by
7.9k points

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