Flexibility -- multiple methods for solutions?

Yh Hoo

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 cant 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 ??

Yh Hoo

this is not a homework but out of my curiosity! anyone please help me.

Diffy

In method 1 you have solved for the square root of x

In method 2 you have solved for x.

Do you see the difference?

DonAntonio

So, after all, you got ONLY one actual real solution to the original equation, didn't you?

Both methods above restrict the possible real solutions, which MUST be non-negative, so in this case in just the same as

the good 'ol age-problems in junior high school, when one had to find out the ages of two people and sometimes

one got a negative solution, which had to be discarded as it didn't fit...

When rooting-squaring, the secret is simple: do whatever you will, but at the end substitute in the original equation

to be sure

DonAntonio

Stephen Tashi

Squaring both sides of an equation may produce an equation which has a larger solution set than the original equation. This comes from the fact that [itex] (\sqrt{x})^2 [/itex] is not always [itex] x [/itex]. If you dealing only with the real numbers then [itex] (\sqrt{-6})^2 [/itex] is not defined while [itex] -6 [/itex] is.

In general [itex] (x^a)^b [/itex] is not always equal to [itex]x^{ab}[/itex].

One of the pleasing powers of mathematics is that it allows us to solve problems by manipulating symbols without any verbal thinking. However, this power always falls slightly short of eliminating the need for verbal thoughts altogether. You have discovered some examples where this is the case. When students are introduced to algebra they often expect to write everything down as lists of symbols and not use words. Some teacher encourage this since it makes papers easier to grade. The truth is that you can't really do algebra in valid manner without writing some words here and there to explain your steps.

DonAntonio

Perhaps one should add that "in general" above refers to complex exponentiation and\or non-positive base (and, thus, again

complex stuff), since indeed [itex]\left(x^a\right)^b=x^{ab}[/itex] whenever the basis is positive and the exponent are real numbers.

DonAntonio