What are my zeros (x^2 - 8x + 1)?

Mathematics

1 answer:

Pavel [41]2 years ago

7 0

Answer:

x= 7.87298334621 x=.12701665379

Step-by-step explanation:

This is in standard form so we have to factor this into

x^2-8x + 16 + 1 = 16

(x-4)^2 = 15

(x-4) = $\sqrt{15\\}$

x-4= ± 3.87298334621

x-4 = 3.87298334621 x-4= -3.87298334621

x= 7.87298334621 x=.12701665379

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Please help! If possible please show your work. Thank you in advance!! :D

Rudiy27

Answer:

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Step-by-step explanation:

The first step in solving the equation is to cube both sides:

(∛x)³ = (-4)³ . . . . . = (-4)(-4)(-4) = 16(-4) = -64

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We're not sure what "checking" is supposed to involve here. Usually, one would check the answer by seeing if a true statement is made when the answer is put into the original equation.

∛(-64) = -4 . . . true

Many calculators will not compute √(-64) because they compute roots using logarithms. The log of a negative number is not defined.

So, the way one would check this is to cube both sides, which is how we got the answer in the first place. We expect the same result from doing the same operation again, so it isn't really a check.

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