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.
+214 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.
