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What are the zero(s) #1x^2-6x+20=0#?

Answer 1

There are no zeros for the function specified.

I first attempted to solve this using the quadratic formula:

#(-b+-sqrt(b^2-4ac))/(2a)#

However, the #4ac# term ends up being much larger than #b^2#, making the term under the radical negative and therefore imaginary.

My next thought was to plot and just check if the graph crosses the x-axis:

graph{x^2-6x+20 [-37.67, 42.33, -6.08, 33.92]}

As you can see, the plot doesn't cross the x-axis, and therefore has no 'zeros'.

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Answer 2

Savannah Anderson

To find the zeros of (1x^2 - 6x + 20 = 0), you can use the quadratic formula:

[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}]

Where (a = 1), (b = -6), and (c = 20). Substituting these values:

[x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4(1)(20)}}}}{{2(1)}}] [= \frac{{6 \pm \sqrt{{36 - 80}}}}{{2}}] [= \frac{{6 \pm \sqrt{{-44}}}}{{2}}]

Since the discriminant ((-44)) is negative, the roots are complex numbers. Therefore, the zeros of the equation (1x^2 - 6x + 20 = 0) are complex numbers and cannot be expressed as real numbers.

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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.