- #1
Yh Hoo
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Flexibility -- multiple methods for solutions?
I found that for some mathematical equations, for example quadratic equations or other equations where f(x)=0, the solutions for f(x)=0 could be 2 i we solve certain way and only 1 out of 2 if we use another method, somethings like changing the nature of the equations.
For example, consider an equation involving surd.
x + 3√x - 18 = 0
Method 1
If i solve this by treating x as (√x)[itex]^{2}[/itex], the new equation would be a quadratic eqt in terms of √x.
(√x)[itex]^{2}[/itex] + 3√x - 18 = 0
∴By applying the quadratic equation formula, √x = 3 or -6 ,where √x=-6 should be ignore right? so we got only 1 solution which is x=9 while -6 is prohibited , meaning can't be substituted even in the original equation!
Method 2
Now if i rearrange the equation so that the term with surd is on one side and without surd is on the other side, we gt
3√x = 18 - x
square both side, we gt
9x = (18 - x)[itex]^{2}[/itex]
x - 45x + 324 = 0
Finally by quadratic equation formula,
x= 9 or 36
but somehow the 36 here is not a solution to the original equation. This is the place i wonder why even it is not a solution we still can get x=36 when f(x)=0 ??
I found that for some mathematical equations, for example quadratic equations or other equations where f(x)=0, the solutions for f(x)=0 could be 2 i we solve certain way and only 1 out of 2 if we use another method, somethings like changing the nature of the equations.
For example, consider an equation involving surd.
x + 3√x - 18 = 0
Method 1
If i solve this by treating x as (√x)[itex]^{2}[/itex], the new equation would be a quadratic eqt in terms of √x.
(√x)[itex]^{2}[/itex] + 3√x - 18 = 0
∴By applying the quadratic equation formula, √x = 3 or -6 ,where √x=-6 should be ignore right? so we got only 1 solution which is x=9 while -6 is prohibited , meaning can't be substituted even in the original equation!
Method 2
Now if i rearrange the equation so that the term with surd is on one side and without surd is on the other side, we gt
3√x = 18 - x
square both side, we gt
9x = (18 - x)[itex]^{2}[/itex]
x - 45x + 324 = 0
Finally by quadratic equation formula,
x= 9 or 36
but somehow the 36 here is not a solution to the original equation. This is the place i wonder why even it is not a solution we still can get x=36 when f(x)=0 ??