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Help solving an equation
- Thread starter CB1101
- Start date
so...\(\displaystyle \frac{x+1}{x}-\frac{x}{x+1}=0 \)
\(\displaystyle \frac{(x+1)^2}{x(x+1)}-\frac{x^2}{x(x+1)}=0 \)
\(\displaystyle \frac{(x+1)^2-x^2}{x(x+1)}=0 \)
Can you continue the exercise?
(x+1)^2 - x^2 = x(x+1)
(x+1)^2 - x^2 = x^2 + x
(x+1)^2 = x
is that it?
No.
Since \(\displaystyle \frac{(x+1)^2-x^2}{x(x+1)}=0 \) , the numerator must be equal to zero, so \(\displaystyle (x+1)^2-x^2 =0 \).
CB1101, what Mathmari means is that the only way that a fraction can = 0 is if the numerator = 0. Does that make sense? I can have the most God awful fraction like this:
\(\displaystyle \dfrac{3x(x-1)(x-4)^3}{x^{e}(4x-3)^5(x^5-7x^3+8)^{\ln9}}=0\)
and all you have to do is set the numerator = 0 and solve. In this case x = 0, 1 and 4.
CB1101, what Mathmari means is that the only way that a fraction can = 0 is if the numerator = 0. Does that make sense? I can have the most God awful fraction like this:
\(\displaystyle \dfrac{3x(x-1)(x-4)^3}{x^{e}(4x-3)^5(x^5-7x^3+8)^{\ln9}}=0\)
and all you have to do is set the numerator = 0 and solve. In this case x = 0, 1 and 4.
I see what you're saying. So it becomes 0/(x(x+1)) = 0
I'm confused as to how to solve when there's a 0 in the numerator
To be preciseI see what you're saying. So it becomes 0/(x(x+1)) = 0
I'm confused as to how to solve when there's a 0 in the numerator
\(\displaystyle \dfrac{f(x)}{g(x)} = 0 \iff f(x) = 0\ and\ g(x) \ne 0.\)
So if you have to solve f(x) / g(x) = 0 all you have to do is to solve f(x) = 0.
Specifically, you must also restrict your x-values to ones that
do not cause the expression (which when it is set equal to zero) to be undefined.
The denominator cannot equal 0, so \(\displaystyle \ x(x + 1) \ne 0 \ \implies\)
that neither \(\displaystyle \ x = 0 \ \ nor \ \ x = -1.\)
do not cause the expression (which when it is set equal to zero) to be undefined.
The denominator cannot equal 0, so \(\displaystyle \ x(x + 1) \ne 0 \ \implies\)
that neither \(\displaystyle \ x = 0 \ \ nor \ \ x = -1.\)