Michael solved the rational equation $$ \frac{x+2}{x}+\frac{2x+1}{x+2}=\frac{−4x+4}{x^2+2x} $$ in the following steps:
(1) He multiplied both sides of the equation by the expression $x(x+2)$: $$ (x+2)^2+x(2x+1)=−4x+4 $$
(2) He squared the binominal and removed the parentheses multiplying by the factor $x$: $$ x^2+4x+4+2x^2+x=−4x+4 $$
(3) Combining like terms he got a quadratic equation and brought it to a factored form: $$\begin{aligned} 3x^2+9x&=0 \cr 3x(x+3)&=0 \end{aligned}$$
(4) A product of factors is zero if and only if one or more of the factors is zero, so Michael concluded that the equation has two solutions, namely $x=0$ and $x=−3$.
Is his solution correct? If not, identify all his wrong steps.
No, his solution is not correct. When solving a rational equation, we have two options: either examine the equation at the beginning to identify values that would result in zero denominators or perform a check at the end.
No, his solution is not correct. In addition to the solutions $x=0$ and $x=−3$ he should have added the solution $x=−2$, which follows immediately from the assignment. (Zero denominator).
Yes. The whole solution is perfectly fine.
No, the error is in step (3). He should have solved the quadratic equation using the formula: $$ x_{1,2}=\frac{−9 \pm \sqrt{9^2−4\cdot 3\cdot 0}}{2} $$ By calculating, he would get the solutions $x=0$ and $x=−9$.
In the original equation, the values $x=0$ and $x=−2$ make denominators equal to zeroes. Since division by zero is undefined, they cannot be solutions to the original equation.
If we do not specify conditions and start solving a rational equation by multiplying both sides of the equation by the LCD (least common denominator), we must always check our solutions to the original equation. We can obtain extraneous solutions. By doing the check for the given equation we find that it has only one solution:
For $x=−3:~L=\frac{−3+2}{−3}+\frac{2\cdot (−3)+1}{−3+2}=\frac{1}{3}+5=\frac{16}{3},~R=\frac{−4\cdot (−3)+4}{(−3)^2+2\cdot (−3)}=\frac{16}{3}.$
For $x=0:~L=\frac{0+2}{0}+\frac{2 \cdot 0+1}{0+2} \mathrm{~(undefined)},~R=\frac{−4\cdot 0+4}{0^2+2 \cdot 0} \mathrm{~(undefined)}$
We can see that $x=0$ is not a solution.