Asked by Unknown"
Check And Help?
1. √x-10+4=0
Answer: x=26
2.√3x+1=7
Answer: x=16
Help With These:
3. √x+6=x
how would you solve this?
I don't get how would It?
4. √x+4 = √x-1+1
it looks similar to problem 2 but it doesn't quite have the same amount of numbers in the same spots..
So I'm stuck on how to solve it.
1. √x-10+4=0
Answer: x=26
2.√3x+1=7
Answer: x=16
Help With These:
3. √x+6=x
how would you solve this?
I don't get how would It?
4. √x+4 = √x-1+1
it looks similar to problem 2 but it doesn't quite have the same amount of numbers in the same spots..
So I'm stuck on how to solve it.
Answers
Answered by
Steve
√(x-10) + 4 = 0
if x=26, you have
√16 + 4 = 0
which is not true.
√n is a positive value, so there is no solution.
Now, √(x-10) - 4 = 0 has solution x=26
√(3x+1) = 7
has the solution x=16 as you say.
√(x+6) = x
square both sides:
x+6 = x^2
now solve as a quadratic
x = -2 or 3
However, √4 = 2, not -2, so -2 is an extraneous solution introduced by squaring. -2+6 = 4 = (-2)^2, but √4 = 2, not -2.
same method here:
x+4 = (x-1) + 2√(x-1) + 1
4 = 2√(x-1)
16 = 4(x-1)
x = 5
check: √(5+4) = √9 = 3 = √(5-1)+1 OK
if x=26, you have
√16 + 4 = 0
which is not true.
√n is a positive value, so there is no solution.
Now, √(x-10) - 4 = 0 has solution x=26
√(3x+1) = 7
has the solution x=16 as you say.
√(x+6) = x
square both sides:
x+6 = x^2
now solve as a quadratic
x = -2 or 3
However, √4 = 2, not -2, so -2 is an extraneous solution introduced by squaring. -2+6 = 4 = (-2)^2, but √4 = 2, not -2.
same method here:
x+4 = (x-1) + 2√(x-1) + 1
4 = 2√(x-1)
16 = 4(x-1)
x = 5
check: √(5+4) = √9 = 3 = √(5-1)+1 OK
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