How do you solve the equation and identify any extraneous solutions for # sqrt(x^2+5)=3x#?
We can square both sides of the equation. But we have to remember that the outcome of a radical is allways
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To solve the equation sqrt(x^2+5)=3x and identify any extraneous solutions, follow these steps:

Square both sides of the equation to eliminate the square root: (sqrt(x^2+5))^2 = (3x)^2.

Simplify both sides: x^2 + 5 = (3x)^2.

Expand the right side: x^2 + 5 = 9  6x + x^2.

Combine like terms: x^2  x^2 + 6x = 9  5.

Simplify further: 6x = 4.

Divide both sides by 6: x = 4/6.

Simplify the fraction: x = 2/3.

Check for extraneous solutions by substituting x = 2/3 back into the original equation.

sqrt((2/3)^2 + 5) = 3  (2/3).

Simplify both sides: sqrt(4/9 + 45/9) = 9/3  2/3.

Simplify further: sqrt(49/9) = 7/3.

Calculate the square root: 7/3 = 7/3.

Since the equation is true, x = 2/3 is a valid solution and there are no extraneous solutions.
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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