Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model.

a√x+b+c=d

Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.

Question

Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model.

a√x+b+c=d

Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.

anonymous · Answer

I need help answering the question

anonymous · Answer

please someone help have to finish flvs

jdoe0001 · Answer

well...  the extraneous would be the one... may take more

the non-extraneous is just any equation more or less

the extraneous  could be $\bf 3\sqrt{x}=-5\implies \sqrt{x}=-\cfrac{5}{3}\implies x=\left( \cfrac{-5}{3} \right)^2\implies x=\cfrac{(-5)^2}{3^2}
\ \quad \
x=\cfrac{25}{9}$

now.. why is that extraneous? or strange or odd ?

well... because positive square root of a number, will never yield a negative root
yet the procedure at $\sqrt{x}=-\cfrac{5}{3}$
assumes that

jdoe0001 · Answer

as far as a non-extraneous, well, just grab a couple of values
get an equation, any will do

anonymous · Answer

thank you I had no idea even where to start until you helped

jdoe0001 · Answer

yw

anonymous · Answer

thx