Let's solve the equation step-by-step:
[tex]\[ 3 x^{\frac{1}{2}} + 5 - 2 x^{-\frac{1}{2}} = 0 \][/tex]
1. Rewrite the equation:
[tex]\[ 3 \sqrt{x} + 5 - 2 \frac{1}{\sqrt{x}} = 0 \][/tex]
2. Introduce a substitution:
Let [tex]\( y = \sqrt{x} \)[/tex]. Hence, [tex]\( x = y^2 \)[/tex] and [tex]\( \frac{1}{\sqrt{x}} = \frac{1}{y} \)[/tex].
3. Rewrite the equation using the substitution:
[tex]\[ 3y + 5 - 2\frac{1}{y} = 0 \][/tex]
4. Clear the fraction by multiplying every term by [tex]\( y \)[/tex]:
[tex]\[ 3y^2 + 5y - 2 = 0 \][/tex]
5. Solve the quadratic equation:
We solve [tex]\( 3y^2 + 5y - 2 = 0 \)[/tex] using the quadratic formula:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 3 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = -2 \)[/tex].
6. Compute the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 5^2 - 4 \cdot 3 \cdot (-2) = 25 + 24 = 49 \][/tex]
7. Substitute into the quadratic formula:
[tex]\[
y = \frac{-5 \pm \sqrt{49}}{2 \cdot 3} = \frac{-5 \pm 7}{6}
\][/tex]
This gives us two solutions:
[tex]\[
y = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3}
\][/tex]
[tex]\[
y = \frac{-5 - 7}{6} = \frac{-12}{6} = -2
\][/tex]
8. Back-substitute to find [tex]\( x \)[/tex]:
Recall [tex]\( y = \sqrt{x} \)[/tex].
For [tex]\( y = \frac{1}{3} \)[/tex]:
[tex]\[
\sqrt{x} = \frac{1}{3} \implies x = \left( \frac{1}{3} \right)^2 = \frac{1}{9}
\][/tex]
For [tex]\( y = -2 \)[/tex]:
[tex]\[
\sqrt{x} = -2
\][/tex]
Since [tex]\( \sqrt{x} \)[/tex] cannot be negative, [tex]\( y = -2 \)[/tex] is not a valid solution.
9. Conclusion:
The valid solution is:
[tex]\[
x = \frac{1}{9}
\][/tex]
Therefore, the solution to the equation [tex]\( 3 x^{\frac{1}{2}} + 5 - 2 x^{-\frac{1}{2}} = 0 \)[/tex] is:
[tex]\[ x = 0.111111111111111 \][/tex] which is [tex]\( \frac{1}{9} \)[/tex].
