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How do you solve #y = x^2 - 6x + 8 = 0# using the quadratic formula?

Answer 1

Avery Alexander

Use the coefficients with the formula, and you will find that #x={2,4}#

The following is how the quadratic formula is written:

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

In this case, the employed quadratic equation looks like this:

#ax^2+bx+c=0#

By applying this pattern, the following is evident:

#a=1# #b=-6# #c=8#

Now let us enter these figures into the formula:

#x=(-(-6)+-sqrt((-6)^2-4(1)(8)))/(2(1))#

#x=(6+-sqrt(36-32))/2#

#x=(6+-sqrt(4))/2=(6+-2)/2=3+-1#

#x={(3-1),(3+1)}#

#color(green)(x={2,4}#

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Answer 2

Hazel Anglin

To solve the quadratic equation (y = x^2 - 6x + 8 = 0) using the quadratic formula, you use the formula:

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

where (a = 1), (b = -6), and (c = 8). Plugging these values into the formula, you get:

[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)} ]

[ x = \frac{6 \pm \sqrt{36 - 32}}{2} ]

[ x = \frac{6 \pm \sqrt{4}}{2} ]

[ x = \frac{6 \pm 2}{2} ]

This gives two solutions:

[ x_1 = \frac{6 + 2}{2} = 4 ]

[ x_2 = \frac{6 - 2}{2} = 2 ]

So, the solutions to the equation are (x = 4) and (x = 2).

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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.