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Question

anonymous · Answer

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. 

Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous. 

Part 3. Explain why the first equation has an extraneous solution and the second does not.

ganeshie8 · Answer

try this for extraneous solution :
$\large  2\sqrt{x+1}+4 = 2 $

try this for non-extraneous solution :
$\large  2\sqrt{x+1}+2 = 4 $

anonymous · Answer

Thank You!

anonymous · Answer

Wait Just A Quick Question why does the first one have an extraneous solution but the second one doesn't?

ganeshie8 · Answer

we will get to know that in partB

ganeshie8 · Answer

solve each equation and see what u get

anonymous · Answer

ok!

anonymous · Answer

no solution for the first?

anonymous · Answer

and the second one's solution is 1/4

anonymous · Answer

right?

ganeshie8 · Answer

try again

anonymous · Answer

brb away for lunch forreal this time

ganeshie8 · Answer

enjoy ur meal :)

anonymous · Answer

my mom is yelling at me to stop doing math (psh that's a first)

anonymous · Answer

thx :)

ganeshie8 · Answer

equation with `extraneous solution` :
$\large 2\sqrt{x+1} + 4 = 2$

$\large 2\sqrt{x+1}  = -2$

$\large \sqrt{x+1}  = -1$

$\large x+ 1 = 1$

$\large x = 0$



lets verify if $x=0$ satisfies the equation :
$\large 2\sqrt{0+1} + 4 = 2$
$\large 2 + 4 =2$
$\large 6 = 2$
FALSE.

So $x=0$ does not satisfy this equaion and hence it is an `extraneous solution` for this equation

ganeshie8 · Answer

similarly you can do the work for solving the second equation

anonymous · Answer

Oh ok Thank You @ganeshie8