solve: $\frac{x^{2}-x - 6}{x^{2}}=\frac{x - 6}{2x}+\frac{2x + 12}{x}$. after multiplying each side of the…

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 $3x^{2}+20x + 12 = 0$. what are the solutions to the equation? $x=-6$ and $x=-\frac{2}{3}$; $x=-6$ and $x=\frac{2}{3}$; $x = 6$ and $x=\frac{2}{3}$

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 $3x^{2}+20x + 12 = 0$. what are the solutions to the equation? $x=-6$ and $x=-\frac{2}{3}$; $x=-6$ and $x=\frac{2}{3}$; $x = 6$ and $x=\frac{2}{3}$

Answer

Explanation:

Step1: Factor the quadratic equation

For the quadratic equation $3x^{2}+20x + 12=0$, we use the AC - method. Here, $a = 3$, $b=20$, $c = 12$, and $ac=3\times12 = 36$. We need to find two numbers that multiply to 36 and add up to 20. The numbers are 2 and 18. So we rewrite the middle - term: $3x^{2}+2x+18x + 12=0$.

Step2: Group and factor by grouping

Group the terms: $(3x^{2}+2x)+(18x + 12)=0$. Factor out the greatest common factor from each group: $x(3x + 2)+6(3x + 2)=0$. Then, $(3x + 2)(x + 6)=0$.

Step3: Apply the zero - product property

If $(3x + 2)(x + 6)=0$, then either $3x+2 = 0$ or $x + 6=0$. For $3x+2 = 0$, we solve for $x$: $3x=-2$, so $x=-\frac{2}{3}$. For $x + 6=0$, we solve for $x$: $x=-6$.

Answer:

$x=-6$ and $x =-\frac{2}{3}$