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madreJ [45]
2 years ago
6

Help hell help help gelp help

Mathematics
1 answer:
IgorC [24]2 years ago
3 0
Just read what it says not that hard go to the left twice and go up 4 units.

Hope you have a good day
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\bf ~~~~~~~~~~~~\textit{negative exponents}
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2 years ago
Which of the following is an extraneous solution of (45-3x)^1/2=x-9?
Iteru [2.4K]

Answer:

C. x =3

Step-by-step explanation:

Extraneous solution is that root of a transformed equation that doesn't satisfy the equation in it's original form because it was excluded from the domain of the original equation.

Let's solve the equation first

\sqrt{45-3x} = x-9\\Taking\ square\ on\ both\ sides\\{(\sqrt{45-3x})}^2 = {(x-9)}^2\\45-3x = x^2-18x+81\\0 = x^2-18x+81-45+3x\\x^2-15x+36 = 0\\x^2-12x-3x+36 = 0\\x(x-12)-3(x-12) = 0\\(x-3)(x-12)\\x-3 = 0\\=> x =3\\x-12 = 0\\x = 12\\We\ will\ check\ the\ solutions\ one\ by\ one\\So,\\for\ x=3\\\sqrt{45-3(3)} = 3-9\\\sqrt{45-9} = -6\\\sqrt{36}= -6\\6\neq -6\\For x=12\\\sqrt{45-3(12)} = 12-9\\\sqrt{45-36} = 6\\\sqrt{36}= 6\\6=6

Hence, we can conclude that x=3 is an extraneous solution of the equation ..

3 0
3 years ago
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