How do you find the roots, real and imaginary, of #y=-2(x+1)^2-(x-3)^2-10 # using the quadratic formula?
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To find the roots of (y = -2(x+1)^2 - (x-3)^2 - 10), we first simplify the equation. Then, we can identify the coefficients and apply the quadratic formula. However, it's worth noting that this equation isn't in standard quadratic form. Instead, we can expand and rearrange it to resemble a quadratic equation before proceeding with the quadratic formula.
[ y = -2(x+1)^2 - (x-3)^2 - 10 ] [ y = -2(x^2 + 2x + 1) - (x^2 - 6x + 9) - 10 ] [ y = -2x^2 - 4x - 2 - x^2 + 6x - 9 - 10 ] [ y = -3x^2 + 2x - 21 ]
Now, we have the equation in standard quadratic form: (y = ax^2 + bx + c) where (a = -3), (b = 2), and (c = -21). We can use the quadratic formula to find the roots:
[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}] [x = \frac{{-2 \pm \sqrt{{2^2 - 4(-3)(-21)}}}}{{2(-3)}}] [x = \frac{{-2 \pm \sqrt{{4 - 252}}}}{{-6}}] [x = \frac{{-2 \pm \sqrt{{-248}}}}{{-6}}] [x = \frac{{-2 \pm 2i\sqrt{62}}}{{-6}}] [x = \frac{{1 \pm i\sqrt{62}}}{{3}}]
So, the roots of the equation are:
[x = \frac{{1 + i\sqrt{62}}}{{3}}] (Imaginary) [x = \frac{{1 - i\sqrt{62}}}{{3}}] (Imaginary)
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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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