Answer :
Certainly! Let's solve the equation step-by-step:
Given the equation:
[tex]\[ 3 \sqrt{x} + 5 - \frac{2}{\sqrt{x}} = 0 \][/tex]
Step 1: Introduce a substitution to simplify the equation. Let [tex]\( y = \sqrt{x} \)[/tex]. This implies [tex]\( x = y^2 \)[/tex].
Step 2: Substitute [tex]\( y \)[/tex] into the equation:
[tex]\[ 3y + 5 - \frac{2}{y} = 0 \][/tex]
Step 3: To consolidate terms, multiply every term by [tex]\( y \)[/tex] to clear the fraction:
[tex]\[ 3y^2 + 5y - 2 = 0 \][/tex]
Step 4: Now we have a quadratic equation in terms of [tex]\( y \)[/tex]. Let's solve the quadratic equation [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].
Step 5: Plug in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{25 + 24}}{6} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{49}}{6} \][/tex]
[tex]\[ y = \frac{-5 \pm 7}{6} \][/tex]
Step 6: This gives us two potential solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
[tex]\[ y = \frac{-5 - 7}{6} = \frac{-12}{6} = -2 \][/tex]
Step 7: Recall that [tex]\( y = \sqrt{x} \)[/tex]. Thus:
[tex]\[ \sqrt{x} = \frac{1}{3} \][/tex]
[tex]\[ \sqrt{x} = -2 \][/tex] (But since [tex]\(\sqrt{x}\)[/tex] must be non-negative, we discard this solution.)
Step 8: Solve for [tex]\( x \)[/tex] using [tex]\( \sqrt{x} = \frac{1}{3} \)[/tex]:
[tex]\[ x = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \][/tex]
So, the final solution is:
[tex]\[ x = \frac{1}{9} \][/tex]
Therefore, the solution to the equation [tex]\( 3 \sqrt{x} + 5 - \frac{2}{\sqrt{x}} = 0 \)[/tex] is:
[tex]\[ x = 0.111111111111111 \][/tex] when expressed as a decimal.
Given the equation:
[tex]\[ 3 \sqrt{x} + 5 - \frac{2}{\sqrt{x}} = 0 \][/tex]
Step 1: Introduce a substitution to simplify the equation. Let [tex]\( y = \sqrt{x} \)[/tex]. This implies [tex]\( x = y^2 \)[/tex].
Step 2: Substitute [tex]\( y \)[/tex] into the equation:
[tex]\[ 3y + 5 - \frac{2}{y} = 0 \][/tex]
Step 3: To consolidate terms, multiply every term by [tex]\( y \)[/tex] to clear the fraction:
[tex]\[ 3y^2 + 5y - 2 = 0 \][/tex]
Step 4: Now we have a quadratic equation in terms of [tex]\( y \)[/tex]. Let's solve the quadratic equation [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].
Step 5: Plug in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{25 + 24}}{6} \][/tex]
[tex]\[ y = \frac{-5 \pm \sqrt{49}}{6} \][/tex]
[tex]\[ y = \frac{-5 \pm 7}{6} \][/tex]
Step 6: This gives us two potential solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-5 + 7}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
[tex]\[ y = \frac{-5 - 7}{6} = \frac{-12}{6} = -2 \][/tex]
Step 7: Recall that [tex]\( y = \sqrt{x} \)[/tex]. Thus:
[tex]\[ \sqrt{x} = \frac{1}{3} \][/tex]
[tex]\[ \sqrt{x} = -2 \][/tex] (But since [tex]\(\sqrt{x}\)[/tex] must be non-negative, we discard this solution.)
Step 8: Solve for [tex]\( x \)[/tex] using [tex]\( \sqrt{x} = \frac{1}{3} \)[/tex]:
[tex]\[ x = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \][/tex]
So, the final solution is:
[tex]\[ x = \frac{1}{9} \][/tex]
Therefore, the solution to the equation [tex]\( 3 \sqrt{x} + 5 - \frac{2}{\sqrt{x}} = 0 \)[/tex] is:
[tex]\[ x = 0.111111111111111 \][/tex] when expressed as a decimal.