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1. Find all real numbers a that satisfy \frac{1}{64a^3 + 7} - 7 = 0.

 

2. Find all values of a that satsify the equation \frac{a}{3} + 1 = \frac{a + 3}{a} + 1.

 May 2, 2023
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1. Find all real numbers a that satisfy\(\frac{1}{64a^3 + 7} - 7 = 0\).

Move the seven to the right side.

\(\frac{1}{64a^3+7}=7\)

Multiply both sides by \(64a^3+7\).

\(1=448a^3+49\)

\(-48=448a^3\)

\(-3=28a^3\)

\(-\frac{3}{28}=a^3\)

\(a=-\sqrt[3]{\frac{3}{28}}\)

There is only one such real number.

2. Find all values of a that satisfy the equation\(\frac{a}{3} + 1 = \frac{a + 3}{a} + 1\).

Remove the 1:

\(\frac{a}{3}= \frac{a + 3}{a}\)

Multiply by 3a:

\(a^2=3a+9\)

\(a^2-3a-9=0\)

Solve using the quadratic formula:

\(\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \left(-9\right)}}{2\cdot \:1}\)

\(\frac{3\pm\sqrt{9-4\cdot-9}}{2}\)

\(\frac{3\pm\sqrt{9--36}}{2}\)

\(\frac{3\pm\sqrt{45}}{2}\)

\(\frac{3\pm3\sqrt{5}}{2}\)

You may choose what to do with that answer.

 May 3, 2023

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