How do you solve #x^2-4x +12=0 #?
See a solution process below:
via Purplemath.com/modules/quadform.htm
According to the quadratic formula,
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To solve the quadratic equation (x^2 - 4x + 12 = 0), you can use the quadratic formula:
[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]
where (a = 1), (b = -4), and (c = 12). Substituting these values into the formula:
[x = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(1)(12)}}}}{{2(1)}}]
[x = \frac{{4 \pm \sqrt{{16 - 48}}}}{{2}}]
[x = \frac{{4 \pm \sqrt{{-32}}}}{{2}}]
[x = \frac{{4 \pm \sqrt{{-1 \cdot 32}}}}{{2}}]
[x = \frac{{4 \pm \sqrt{{-1}} \cdot \sqrt{{32}}}}{{2}}]
[x = \frac{{4 \pm 4i \sqrt{{2}}}}{{2}}]
[x = \frac{{4}}{{2}} \pm \frac{{4i \sqrt{2}}}{{2}}]
[x = 2 \pm 2i\sqrt{2}]
Therefore, the solutions to the equation (x^2 - 4x + 12 = 0) are (x = 2 + 2i\sqrt{2}) and (x = 2 - 2i\sqrt{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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