Question

solve: \\(\frac{x^2 - x - 6}{x^2} = \frac{x - 6}{2x} + \frac{2x + 12}{x}\\) after multiplying each side of the equation by the lcd and simplifying, the resulting equation is \\(3x^2 - 20x + 12 = 0\\) \\(3x^2 - 20x - 12 = 0\\) \\(3x^2 + 20x + 12 = 0\\) \\(3x^2 + 20x - 12 = 0\\)

Explanation:

Step1: Find the LCD

The denominators are \(x^2\), \(2x\), and \(x\). The least common denominator (LCD) is \(2x^2\) (since we take the highest power of \(x\) which is \(x^2\) and the highest coefficient for the constant factor, here 2 from \(2x\)).

Step2: Multiply each term by LCD

Multiply each term in the equation \(\frac{x^2 - x - 6}{x^2}=\frac{x - 6}{2x}+\frac{2x + 12}{x}\) by \(2x^2\):

• For the left - hand side: \(\frac{x^2 - x - 6}{x^2}\times2x^2=2(x^2 - x - 6)=2x^2-2x - 12\)
• For the first term on the right - hand side: \(\frac{x - 6}{2x}\times2x^2=x(x - 6)=x^2-6x\)
• For the second term on the right - hand side: \(\frac{2x + 12}{x}\times2x^2=2x(2x + 12)=4x^2+24x\)

Step3: Simplify the equation

Now, set up the equation after multiplication:
\(2x^2-2x - 12=x^2-6x + 4x^2+24x\)

Combine like terms on the right - hand side: \(x^2+4x^2=5x^2\) and \(-6x + 24x = 18x\), so the right - hand side is \(5x^2+18x\)

The equation becomes \(2x^2-2x - 12=5x^2+18x\)

Subtract \(2x^2-2x - 12\) from both sides to get:
\(0 = 5x^2+18x-(2x^2-2x - 12)\)
\(0=5x^2+18x - 2x^2 + 2x+12\)
\(0 = 3x^2+20x + 12\)
Or \(3x^2+20x + 12=0\) (Wait, this is wrong. Let's re - do the multiplication step correctly.)

Wait, let's re - calculate the multiplication:

The original equation: \(\frac{x^{2}-x - 6}{x^{2}}=\frac{x - 6}{2x}+\frac{2x + 12}{x}\)

Multiply each term by \(2x^{2}\):

Left term: \(\frac{x^{2}-x - 6}{x^{2}}\times2x^{2}=2(x^{2}-x - 6)=2x^{2}-2x-12\)

First right term: \(\frac{x - 6}{2x}\times2x^{2}=x(x - 6)=x^{2}-6x\)

Second right term: \(\frac{2x + 12}{x}\times2x^{2}=2x(2x + 12)=4x^{2}+24x\)

Now, the equation is \(2x^{2}-2x - 12=x^{2}-6x+4x^{2}+24x\)

Right side: \(x^{2}+4x^{2}=5x^{2}\), \(-6x + 24x = 18x\), so right side is \(5x^{2}+18x\)

Bring all terms to one side: \(2x^{2}-2x - 12-5x^{2}-18x = 0\)

Combine like terms: \(2x^{2}-5x^{2}=-3x^{2}\), \(-2x-18x=-20x\)

So the equation is \(-3x^{2}-20x - 12 = 0\)

Multiply both sides by - 1: \(3x^{2}+20x + 12 = 0\) (Wait, no, let's check again. Wait, when we move terms from right to left:

\(2x^{2}-2x - 12-(x^{2}-6x)-(4x^{2}+24x)=0\)

\(2x^{2}-2x - 12 - x^{2}+6x-4x^{2}-24x = 0\)

\((2x^{2}-x^{2}-4x^{2})+(-2x + 6x-24x)-12 = 0\)

\((-3x^{2})+(-20x)-12 = 0\)

Multiply both sides by - 1: \(3x^{2}+20x + 12 = 0\)

Wait, there is a mistake in the previous step. Let's do it again:

Original equation: \(\frac{x^{2}-x - 6}{x^{2}}=\frac{x - 6}{2x}+\frac{2x + 12}{x}\)

LCD is \(2x^{2}\)

Multiply each term:

Left: \(2(x^{2}-x - 6)=2x^{2}-2x - 6\times2=2x^{2}-2x - 12\)

First right term: \(x(x - 6)=x^{2}-6x\)

Second right term: \(2x(2x + 12)=4x^{2}+24x\)

Now, the equation is \(2x^{2}-2x - 12=x^{2}-6x + 4x^{2}+24x\)

Combine right - hand side: \(x^{2}+4x^{2}=5x^{2}\), \(-6x + 24x = 18x\), so right - hand side is \(5x^{2}+18x\)

Subtract \(2x^{2}-2x - 12\) from both sides:

\(0=5x^{2}+18x-(2x^{2}-2x - 12)\)

\(0 = 5x^{2}+18x-2x^{2}+2x + 12\)

\(0=3x^{2}+20x + 12\)

So the resulting equation is \(3x^{2}+20x + 12 = 0\)

Answer:

\(3x^{2}+20x + 12 = 0\) (The option with this equation, i.e., the option "3\(x^{2}\)+20\(x\)+12 = 0")