## Algebra: A Combined Approach (4th Edition)

$x=1\pm\sqrt{3}$
$\dfrac{6}{x^{2}}=\dfrac{3}{x+1}$ Multiply the whole equation by $x^{2}(x+1)$ $x^{2}(x+1)\Big(\dfrac{6}{x^{2}}=\dfrac{3}{x+1}\Big)$ $6(x+1)=3x^{2}$ Evaluate the product on the left side: $6x+6=3x^{2}$ Take all terms to the right side of the equation and simplify it by taking out common factor $3$: $0=3x^{2}-6x-6$ $3(x^{2}-2x-2)=0$ Take the $3$ to divide the right side of the equation: $x^{2}-2x-2=\dfrac{0}{3}$ $x^{2}-2x-2=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=1$, $b=-2$ and $c=-2$. Substitute the known values into the formula and simplify: $x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-2)}}{2(1)}=\dfrac{2\pm\sqrt{4+8}}{2}=...$ $...=\dfrac{2\pm\sqrt{12}}{2}=\dfrac{2\pm2\sqrt{3}}{2}=1\pm\sqrt{3}$ The original equation is not undefined for neither of the solutions found. Our final answers are $x=1\pm\sqrt{3}$