How do you solve #0 = x^2 - 5x + 6# using the quadratic formula?
Rewrite in standard form
Use Quadratic formula
Now,
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To solve (0 = x^2 - 5x + 6) using the quadratic formula, which is (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) for the quadratic equation (ax^2 + bx + c = 0), first identify (a), (b), and (c) from the given equation.
In this case, (a = 1), (b = -5), and (c = 6).
Now, substitute these values into the formula:
(x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1})
(x = \frac{5 \pm \sqrt{25 - 24}}{2})
(x = \frac{5 \pm \sqrt{1}}{2})
Since the square root of 1 is 1, this simplifies to:
(x = \frac{5 \pm 1}{2})
So, the two solutions are:
(x = \frac{5 + 1}{2} = \frac{6}{2} = 3)
and
(x = \frac{5 - 1}{2} = \frac{4}{2} = 2)
Therefore, the solutions to (0 = x^2 - 5x + 6) are (x = 3) and (x = 2).
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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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