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!
quadratic/exponential equations
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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.sdilmanian wrote:
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 ?
I would substitute √x=y.2.
How would you solve (factor) this:
x + √x - 12 = 0
Thanks BTG crew!
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.