How do you find the zeros, real and imaginary, of #y= 3x^2-6x+2 # using the quadratic formula?
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To find the zeros of the quadratic function ( y = 3x^2 - 6x + 2 ) using the quadratic formula:
- Identify the coefficients: ( a = 3 ), ( b = -6 ), and ( c = 2 ).
- Apply the quadratic formula: ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ).
- Substitute the coefficients into the formula: ( x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 3 \cdot 2}}}}{{2 \cdot 3}} ).
- Simplify the expression under the square root: ( x = \frac{{6 \pm \sqrt{{36 - 24}}}}{{6}} ).
- Further simplify: ( x = \frac{{6 \pm \sqrt{{12}}}}{{6}} ).
- Simplify the square root: ( x = \frac{{6 \pm 2\sqrt{3}}}{{6}} ).
- Reduce the fraction: ( x = \frac{{3 \pm \sqrt{3}}}{{3}} ).
- The solutions are ( x = \frac{{3 + \sqrt{3}}}{{3}} ) and ( x = \frac{{3 - \sqrt{3}}}{{3}} ), which are the real and imaginary zeros, respectively.
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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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