How do you solve # (x^2 - 5)/(x + 3) = 0#?

Answer 1

I found: #x=+-sqrt(5)#

We need to find the value(s) of #x# that satisfy our equation; we also need to avoid #x# values that make our relationship INDETERMINATE! This is because if you make the denominator zero (choosing #x=-3#) the division by zero is not possible! So as first consideration we say that: #x!=-3# Next we solve our equation:
#x^2-5=0*(x+3)# moving the denominator to the right of the equal sign; so: #x^2-5=0# #x^2=5# #x=+-sqrt(5)#
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Answer 2

To solve the equation (x^2 - 5)/(x + 3) = 0, we set the numerator equal to zero and solve for x.

x^2 - 5 = 0

Adding 5 to both sides:

x^2 = 5

Taking the square root of both sides:

x = ±√5

Therefore, the solutions to the equation are x = √5 and x = -√5.

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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.

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.

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.

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.

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