I have an equation to solve for y:
$\frac{y^2}{y}=1$ Normally, I would cancel out one $y$ and get $y=1$ as a single solution.
But If I think of it as quadratic equation
$y^2=y$ $y^2-y=0$ $y(y-1)=0$ $y=0 \space \text{or} \space y=1$ to have two solutions.
But when I put $y=0$ in original equation, I get $\frac{0}{0}$, so is $y=0$ a solution or not ?
If yes, then I get $\frac{0}{0}$.
If no, then how come this quadratic equation has $1$ solution ?
Quadratic equation : Two solutions or one solution?
1 Answers
This equation has two solutions, $y=0$ and $y=1$:
$y^2=y \tag{1}$
This equation has one solution, $y=1$:
$\frac{y^2}{y} = 1 \tag{2}$
The reason that it has one solution is that either of the operations "Cancel a $y$ from the top and bottom of the fraction" and "Multiply both sides of the equation by $y$" are only valid when $y\neq 0$. In particular, if you want to manipulate (2) to look like (1), you first have to assume that $y\neq 0$, which rules out one of the solutions of the quadratic equation.
A more sophisticated answer realises that when we are asked to "solve" an equation, what we are really doing is looking for a root of a particular function (i.e. a value of the argument at which evaluating the function gives zero). In the case of (1), the function is $f(y)=y^2-y$, and in the case of (2) the function is $f(y)=y^2/y-1$.
Now, functions have to have a domain. In the case of (1) the domain is $\mathbb{R}$ (the real numbers) which includes both 0 and 1. In the case of (2), the domain is $\mathbb{R}-\{0\}$ (the real numbers without 0) which doesn't include 0.
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0OK. Thanks a lot. – 2012-06-28