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A. 6 is a solution to the original equation. The value –5 is an extraneous solution. B. 6 and 5 are both extraneous solutions. C. 5 is a solution to the original equation. The value –6 is an extraneous solution. D. 6 and –5 are solutions.
do u know the meaning of extraneous
That it's unrelated, I think.
square both sides x^2 = x+ 30 x^2 -x - 30 = 0 can u solve this?
extraneous means that although the working out gives it as a solution , it is not possible as a root
5 and 6?
not quite (x - 6)(x + 5) = 0 giving x = 6 , -5
So what do I do with that?
well see if these roots are possible by substituting each into the original equation
eg x = 6 6 = sqrt(6 + 30) 6 = 6 - so that is a genuine root
So I can eliminate B and C
is -5 a solution?
I dont think so.
try plugging in -5 for x in the original equation
- 5 = sqrt (-5 + 30) does that work out?
what are the square roots of 25?
5 ANd -5 right?
so its a genuine root
So they're both genuine making them both solutions.
I do believe I'd pick D
Thank you very much.
please explain why thats wrong
ask the teacher, I answered D and I got it wrong.
is this your question?
oh sorry this is your question |dw:1377219778699:dw|
your answers are 6 and -5!
The right answer should be 6 is a solution to the original equation. The value –5 is an extraneous solution.