How do you find the roots of #x^2-6x+12=0#?

Answer 1

#x=3+-sqrt(3)color(white)(..)i#

Given: #0=x^2-6x+12#

#color(brown)("The question specifically equates to 0. So we must find a solution that")##color(brown)("works for "y=0. #

Consider #y=ax^2+bx+c color(white)("ddd")and color(white)("ddd")x=(-b+-sqrt(b^2-4ac))/(2a)#

Now lets look at the equation part : #b^2-4ac#. THis is called the determinant.

#b^2-4ac color(white)("ddd")->color(white)("d") color(white)("ddd") (-6)^2-4(1)(12) = -12#

As this is negative the graph does not have any x-intercepts where #x# is in the set of what is called real numbers #x !in RR#

There will be a solution of #x# where it is in the set of 'Complex Numbers. #x in CC#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#x=(-b+-sqrt(b^2-4ac))/(2a) color(white)("d")->color(white)("d")x=(+6+-sqrt(-12))/2#

#x=3+-sqrt((-12)/4)#

#x=3+-sqrt(3xx(-1))#

#x=3+-sqrt(3)xxsqrt(-1)#

But #sqrt(-1)=i#

#x=3+-sqrt(3)color(white)(..)i#

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Answer 2

To find the roots of (x^2 - 6x + 12 = 0), you can use the quadratic formula, which is (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a), (b), and (c) are the coefficients of the quadratic equation. For this equation, (a = 1), (b = -6), and (c = 12). Substituting these values into the quadratic formula:

[x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 1 \cdot 12}}}}{{2 \cdot 1}}]

[x = \frac{{6 \pm \sqrt{{36 - 48}}}}{2}]

[x = \frac{{6 \pm \sqrt{{-12}}}}{2}]

Since the square root of -12 is imaginary, the roots are complex. Therefore, the roots are:

[x = \frac{{6 + \sqrt{{-12}}}}{2} = \frac{{6 + 2i\sqrt{3}}}{2} = 3 + i\sqrt{3}]

[x = \frac{{6 - \sqrt{{-12}}}}{2} = \frac{{6 - 2i\sqrt{3}}}{2} = 3 - i\sqrt{3}]

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Answer 3

You can use the quadratic formula to find the roots of the equation (x^2 - 6x + 12 = 0).

The quadratic formula is: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a), (b), and (c) are the coefficients of the quadratic equation (ax^2 + bx + c = 0).

For the given equation (x^2 - 6x + 12 = 0):

  • (a = 1),
  • (b = -6),
  • (c = 12).

Substituting these values into the quadratic formula:

(x = \frac{{-(-6) \pm \sqrt{{(-6)^2 - 4 \cdot 1 \cdot 12}}}}{{2 \cdot 1}})

(x = \frac{{6 \pm \sqrt{{36 - 48}}}}{2})

(x = \frac{{6 \pm \sqrt{{-12}}}}{2})

Since the discriminant (b^2 - 4ac) is negative, the roots will be complex numbers.

(x = \frac{{6 \pm \sqrt{{-12}}}}{2})

(x = \frac{{6 \pm 2i\sqrt{3}}}{2})

(x = 3 \pm i\sqrt{3})

So, the roots of the equation (x^2 - 6x + 12 = 0) are (3 + i\sqrt{3}) and (3 - i\sqrt{3}).

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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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