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Solve the equation identify extraneous solutions

Solve the equation identify extraneous solutions-example-1
User Exceen
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1 Answer

7 votes

Answer:

Explanation:

The goal to solving any equation is to have x = {something}. That means we need to get the x out from underneath that radical. It's a square root, so we can "undo" it by squaring. Square both sides because this is an equation. Squaring both sides gives you


x^2=-3x+40

Get everything on one side of the equals sign and set the quadratic equal to 0:


x^2+3x-40=0

Throw this into the quadratic formula to get that the solutions are x = 5 and -8. We need to see if only one works, both work, or neither work in the original equation.

Does
5=√(-3(5)+40)?


5=√(-15+40) and


5=√(25)

and 5 = 5. So 5 works. Let's try -8 now:


-8=√(-3(-8)+40) and


-8=√(24+40) so


-8=√(64)

-8 = 8? No it doesn't. So only 5 works. Your choice is the third one down.

User Zhibin
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