Solve each equation. Check for extraneous solutions. 1)$$\sqrt[3]{x-3}=1$$ 2) $$\sqrt{x+7}=x+1$$ 3) $$\sqrt{3x-8}=2$$ What are extraneous solutions? Solve and show steps please, this is new to me .

Question

Solve each equation. Check for extraneous solutions.
1)$$\sqrt[3]{x-3}=1$$
2) $$\sqrt{x+7}=x+1$$
3) $$\sqrt{3x-8}=2$$

What are extraneous solutions? Solve and show steps please, this is new to me >.<

bahrom7893 · Answer

square both sides in each equation and then solve

amistre64 · Answer

when trying to make an equation solvable, we introduce false solutions

amistre64 · Answer

the dbl chk weeds out the false solutions

anonymous · Answer

how would you square if it's the cube root

amistre64 · Answer

put a ^2 to it :)

amistre64 · Answer

cube each side ....

anonymous · Answer

can you do the first one and show the steps x_x

amistre64 · Answer

when does cbrt(n) = 1?

anonymous · Answer

when its negative?

amistre64 · Answer

lets try that out;  cbrt(-8) = -2

nope

amistre64 · Answer

to undo a radical you put a power to it

bahrom7893 · Answer

dude come on first one is obviously 4...

amistre64 · Answer

cbrt^3 undo each other

sqrt^2 undo each other

6rt^6 undo each other

amistre64 · Answer

whatever you do to one side, you do to the other as well

anonymous · Answer

i don't understand how to get rid of the cube root.

amistre64 · Answer

$$(\sqrt[n]{\ a\ })^n=a$$

amistre64 · Answer

$$(\sqrt[3]{\ x\ })^3=x$$

anonymous · Answer

ok so it would be x^3-9=3

bahrom7893 · Answer

$$\sqrt[3]{x-3}=1$$$$(\sqrt[3]{x-3})^3=(1)^3$$Do you get it?

amistre64 · Answer

you realy should go back a few steps in math to catch up to where you wanna be now...

bahrom7893 · Answer

good idea.. so should i btw

bahrom7893 · Answer

Quote:
can someone help @amistre64 @myininaya @lalaly @bahrom7893
btw I'm flattered to see my name among the legends. Even though it's the last one :D!

anonymous · Answer

so the answers 1? I'm confused now

anonymous · Answer

honestly i'm just trying to get my homework done, i dont understand this though -.-

bahrom7893 · Answer

I don't see how this is so hard, man seriously u need to get some real tutoring help:
1)$$\sqrt[3]{x-3}=1$$$$(\sqrt[3]{x-3})^3=(1)^3$$$$x-3=1$$$$x=1+3=4$$

anonymous · Answer

ok so for the second one squaring takes out the radical but makes the other part squared?

bahrom7893 · Answer

1) $$\sqrt{x+7}=x+1$$$$(\sqrt{x+7})^2=(x+1)^2$$$$x+7=x^2+2x+1$$Now u can just simplify, factor etc..

bahrom7893 · Answer

after squaring both sides for the last one u'll get:
3x-8=4
3x=12
x=4

anonymous · Answer

Can you help me with 3 more of the same type please, i should of posted these instead because they look harder

anonymous · Answer

$$(2x+1)^{1/3}=3$$

anonymous · Answer

$$\sqrt{x^2-5}=4$$
$$3(x+1)^{4/3}=48$$

anonymous · Answer

-> 2x + 1 = 3^3 
-> 2x = 27 - 1
=>   x = 26/2 = 13

anonymous · Answer

what about the 1/3 power?

anonymous · Answer

x^2 - 5 = 4^2
->x^2 = 16 + 5 = 21
=> x   =  +/- sqrt ( 21)

anonymous · Answer

You always power both sides!

anonymous · Answer

oh so 1/3 power is the same as being cubed ok

anonymous · Answer

3(x+1)^4/3   = 48
->(x+1)^4/3 = 48/ 3 = 16
=> x + 1 = 16 ^ (3/4)
=> x = 8 - 1 = 7

anonymous · Answer

Even you have the result, but you need to plug it back to check for extraneous!

anonymous · Answer

what does extraneous mean & you check for it

anonymous · Answer

how do*

anonymous · Answer

Just plug the root back to the origional see if they match, the equation is true!