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eimsori [14]
2 years ago
8

Which value is NOT a solution of 8x3 – 1 = 0?

Mathematics
1 answer:
Tpy6a [65]2 years ago
7 0

<u><em>Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.</em></u>

Answer:

Any value other than the values x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4} will not be a solution of 8x^3\:-\:1\:=\:0.

Step-by-step explanation:

Considering the equation

8x^3\:-\:1\:=\:0

Steps to solve the equation

8x^3-1=0

\mathrm{Add\:}1\mathrm{\:to\:both\:sides}

8x^3-1+1=0+1

\mathrm{Simplify}

x^3=\frac{1}{8}

\mathrm{Divide\:both\:sides\:by\:}8

\frac{8x^3}{8}=\frac{1}{8}

\mathrm{Simplify}

x^3=\frac{1}{8}

As

\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}

x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}

So,

x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}

Therefore,

Any value other than the values x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4} will not be a solution of 8x^3\:-\:1\:=\:0.

Keywords: solution, value

Learn more about equation solution from  brainly.com/question/1679491

#learnwithBrainly

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And then you add 12, which is 32.

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Step-by-step explanation:

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