Question

@jim_thompson5910

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

Answer 1

plug in y = 0 and solve for x

Answer 2

i got 1735=x^3 and then the cube root closest to that is 12^3 which is 1728

Answer 3

y = (x –12)^3 – 7

0 = (x –12)^3 – 7

7 = (x-12)^3

Keep going until you've isolated x

Answer 4

i get the same thing... im so confused

Answer 5

\[\Large 7 = (x-12)^3\]

\[\Large \sqrt[3]{7} = x-12\]

\[\Large \sqrt[3]{7}+12 = x\]

\[\Large x = \sqrt[3]{7}+12 \]

Answer 6

so then what are the zero's ?

Answer 7

i just showed you

Answer 8

that's a zero because if you plug that in for x, you'll get y = 0

Answer 9

so those are the only two since it will be plus or minus

Answer 10

no there's only one real root..the other two are imaginary