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How do you solve #sqrt[5x-6]=2# and find any extraneous solutions?

Answer 1

Eva Adamson

#x=2#

#color(blue)"square both sides"#

#"note that "sqrtaxxsqrta=(sqrta)^2=a#

#(sqrt(5x-6))^2=2^2#

#rArr5x-6=4#

#"add 6 to both sides"#

#5xcancel(-6)cancel(+6)=4+6#

#rArr5x=10#

#"divide both sides by 5"#

#(cancel(5) x)/cancel(5)=10/5#

#rArrx=2#

#color(blue)"As a check"#

If this value is equal to the right side of the equation after being substituted into the left side, the solution exists.

#sqrt(10-6)=sqrt4=2=" right side"#

#rArrx=2" is the solution"#

#"There are no extraneous solutions"#

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Answer 2

Jameson Allen

To solve the equation sqrt[5x-6]=2, we need to isolate the variable x.

First, square both sides of the equation to eliminate the square root: 5x-6 = 4.

Next, add 6 to both sides of the equation: 5x = 10.

Then, divide both sides by 5 to solve for x: x = 2.

To check for extraneous solutions, substitute the value of x back into the original equation: sqrt[5(2)-6] = 2.

Simplifying, we get sqrt[4] = 2, which is true.

Therefore, the solution x = 2 is valid and there are no extraneous solutions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.