Question

1. when the equation $2x^2 - 40b^7 = 0$ is solved for $x$ in terms of $b$, where $b \geq 0$, then $x =$

(1) $\pm 20b^4\sqrt{b}$
(3) $\pm 2b^3\sqrt{5b}$
(2) $\pm 4b^3\sqrt{2b}$
(4) $\pm 2b^2\sqrt{10b^3}$

Explanation:

Step1: Isolate \(x^2\) term

Start with the equation \(2x^2 - 40b^7 = 0\). Add \(40b^7\) to both sides:
\(2x^2 = 40b^7\)
Divide both sides by 2:
\(x^2 = 20b^7\)

Step2: Simplify \(b^7\) and factor 20

Rewrite \(b^7\) as \(b^{6} \cdot b=(b^3)^2 \cdot b\) and 20 as \(4 \cdot 5\). So:
\(x^2 = 4 \cdot 5 \cdot (b^3)^2 \cdot b\)

Step3: Take square root

Take the square root of both sides. Remember that \(\sqrt{ab}=\sqrt{a}\sqrt{b}\) and \(\sqrt{a^2}=|a|\), but since \(b\geq0\), we can write:
\(x = \pm\sqrt{4 \cdot 5 \cdot (b^3)^2 \cdot b}=\pm\sqrt{4}\cdot\sqrt{(b^3)^2}\cdot\sqrt{5b}\)
Simplify \(\sqrt{4} = 2\) and \(\sqrt{(b^3)^2}=b^3\) (because \(b\geq0\)):
\(x=\pm 2b^3\sqrt{5b}\)

Answer:

(3) \(\pm2b^3\sqrt{5b}\)