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Hi! I need some assistance with these two problems. Thanks in advance!

 

1)

 

2)

 Sep 21, 2019
 #1
avatar+2417 
+2

2. Hint: Add 3 to both sides, then square both sides.

 

Subsituting 0 will get a complex number

1.) Here is an example

 

x=2 is has potential for extraneous roots because if you square both sides, and get x2=4, there are other possible solutions that DON'T solve the original equation.

 

Based on that, what do you think the answer is?

 Sep 21, 2019
edited by CalculatorUser  Sep 21, 2019
edited by CalculatorUser  Sep 21, 2019
 #2
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0

Probably plus&minus 3√2?

Guest Sep 21, 2019
 #3
avatar+2417 
+2

I think so

 

Are you talking algebra 1 or algebra 2 trig honors

CalculatorUser  Sep 21, 2019
 #4
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PreCalc :)

Guest Sep 21, 2019
 #5
avatar+2417 
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yes ok no wonder why it was kind of hard XD I am in 8th grade

CalculatorUser  Sep 21, 2019
 #6
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+2

Solve for x:
sqrt(x^2 - 9) - 3 = 0

Isolate terms with x to the left hand side.
Add 3 to both sides:
sqrt(x^2 - 9) = 3

Eliminate the square root on the left hand side.
Raise both sides to the power of two:
x^2 - 9 = 9

Isolate terms with x to the left hand side.
Add 9 to both sides:
x^2 = 18

Eliminate the exponent on the left hand side.
Take the square root of both sides:

x = 3 sqrt(2)    or     x = -3 sqrt(2)

 Sep 21, 2019

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