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solve the equation. \\(\\frac{3}{x^2} - \\frac{44}{x} = 15\\) select th…

Question

solve the equation.
\\(\frac{3}{x^2} - \frac{44}{x} = 15\\)

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the solution set is {}.
(simplify your answer. type an integer or a fraction. use a comma to separate answers as needed.)

b. the solution is the empty set.

Explanation:

Step1: Eliminate denominators

Multiply each term by \(x^2\) (assuming \(x
eq0\)) to get \(3 - 44x=15x^2\).

Step2: Rearrange into standard quadratic form

Rearrange the equation to \(15x^2 + 44x - 3 = 0\).

Step3: Use quadratic formula

For a quadratic equation \(ax^2+bx+c = 0\), the solutions are \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 15\), \(b = 44\), \(c=-3\).
First, calculate the discriminant \(\Delta=b^2 - 4ac=(44)^2-4\times15\times(-3)=1936 + 180 = 2116\).
Then, \(x=\frac{-44\pm\sqrt{2116}}{2\times15}=\frac{-44\pm46}{30}\).

Step4: Find the two solutions

For the plus sign: \(x=\frac{-44 + 46}{30}=\frac{2}{30}=\frac{1}{15}\).
For the minus sign: \(x=\frac{-44-46}{30}=\frac{-90}{30}=-3\).
We need to check if these solutions make the original equation's denominators non - zero. For \(x=\frac{1}{15}\), \(x^2=\frac{1}{225}
eq0\) and \(x=\frac{1}{15}
eq0\). For \(x = - 3\), \(x^2 = 9
eq0\) and \(x=-3
eq0\). So both solutions are valid.

Answer:

The solution set is \(\{-3,\frac{1}{15}\}\)