How do you find the roots, real and imaginary, of #y=-2x^2 + 7x +6 # using the quadratic formula?
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To find the roots of (y = -2x^2 + 7x + 6) using the quadratic formula, you first identify the coefficients: (a = -2), (b = 7), and (c = 6). Then, you apply the quadratic formula:
[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]
Substitute the coefficients into the formula:
[x = \frac{{-7 \pm \sqrt{{7^2 - 4(-2)(6)}}}}{{2(-2)}}]
[x = \frac{{-7 \pm \sqrt{{49 + 48}}}}{{-4}}]
[x = \frac{{-7 \pm \sqrt{{97}}}}{{-4}}]
Thus, the roots are:
[x = \frac{{-7 + \sqrt{{97}}}}{{-4}}] (real)
[x = \frac{{-7 - \sqrt{{97}}}}{{-4}}] (real)
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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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