quadratic/exponential equations

[sdilmanian]

Hi,

1.
(x-3)(x+4)
---------- [divided by] = 0
(x+2)(x-3)

Since(x-3) cannot possibly equal 0, is it possible to cancel the (x-3) in the numerator and denominator, leaving us with only one solution: x= -4?
Or must we leave both expressions in the numerator, giving us two possible solutions: x= -4 or x=3 ?

2.
How would you solve (factor) this:
x + √x - 12 = 0

Thanks BTG crew!

[shankar.ashwin]

1.
(x-3)(x+4)
---------- [divided by] = 0
(x+2)(x-3)

Since(x-3) cannot possibly equal 0, is it possible to cancel the (x-3) in the numerator and denominator, leaving us with only one solution: x= -4?
Or must we leave both expressions in the numerator, giving us two possible solutions: x= -4 or x=3 ?

Here if x-3=0, we would get a 0/0 which would not be equal to 0. Hence x can take only -4 for it to satisfy the equation.

2.
How would you solve (factor) this:
x + √x - 12 = 0

Thanks BTG crew!

I would substitute √x=y.

Hence the eq. becomes y^2+y-12=0
(y+4)(y-3)=0

y=-4 (or) y=3.

Now, √x=y =-4 (or) x = 16

and, √x=y=3 (or) x =9.

In both cases when we take √x, you get +/- 4 (or) +/-3. But only one value would satisfy the equation. I don't think we can generalize this expression as such.