How do you solve $4 x^{\frac{3}{2}} - 5 = 103$ $4x^(3/2) - 5 = 103$ and find any extraneous solutions?

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Answer 1 · 2016-05-31T21:20:43

$x = 2 . \sqrt[3]{99}$ $x=2color(white)(.)root(3)(99)$

Not sure where the extraneous solutions come from!

Add 5 to both sides

$4 x^{\frac{3}{2}} = 108$ $4x^(3/2)=108$

Write as : $4 \sqrt{x^{3}} = 108$ $4sqrt(x^3)=108$

Divide both sides by 4

$\sqrt{x^{3}} = 27$ $sqrt(x^3)=27$

square both sides

$x^{3} = 792$ $x^3=792" "$

cube root both sides

$x = \sqrt[3]{792} = \sqrt[3]{2^{3} \times 3^{2} \times 11} \leftarrow prime factors$ $x=root(3)(792) = root(3)(2^3xx3^2xx11)" "larr" prime factors"$

$x = 2 . \sqrt[3]{99}$ $x=2color(white)(.)root(3)(99)$

How do you solve 4x^(3/2) - 5 = 1034x32−5=1034x^(3/2) - 5 = 103 and find any extraneous solutions?