How do you find the zeros, real and imaginary, of #y= 3x^2-6x+2 # using the quadratic formula?

Answer 1

#1 +- sqrt3/3#

Use the improved quadratic formula (Google, Yahoo Search) #D = d^2 = b^2 - 4ac = 36 - 24 = 12 --> d = +- 2sqrt3# There are 2 real roots: #x = -b/(2a) +- d/(2a) = 6/6 +- (2sqrt3)/6 = 1 +- sqrt3/3#
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Answer 2

To find the zeros of the quadratic function ( y = 3x^2 - 6x + 2 ) using the quadratic formula:

  1. Identify the coefficients: ( a = 3 ), ( b = -6 ), and ( c = 2 ).
  2. Apply the quadratic formula: ( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} ).
  3. Substitute the coefficients into the formula: ( x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 3 \cdot 2}}}}{{2 \cdot 3}} ).
  4. Simplify the expression under the square root: ( x = \frac{{6 \pm \sqrt{{36 - 24}}}}{{6}} ).
  5. Further simplify: ( x = \frac{{6 \pm \sqrt{{12}}}}{{6}} ).
  6. Simplify the square root: ( x = \frac{{6 \pm 2\sqrt{3}}}{{6}} ).
  7. Reduce the fraction: ( x = \frac{{3 \pm \sqrt{3}}}{{3}} ).
  8. 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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Answer from HIX Tutor

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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