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How do you find the roots, real and imaginary, of #y= x^2 - 5x + 6 # using the quadratic formula?

Answer 1

Eli Assouline

Use factoring ...

#y=(x-2)(x-3)#

roots : #x=2 " and "x= 3#

Hope that helps

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Answer 2

Kennedy Adams

To find the roots of ( y = x^2 - 5x + 6 ) using the quadratic formula, first identify the coefficients:

( a = 1 ) (coefficient of ( x^2 )), ( b = -5 ) (coefficient of ( x )), ( c = 6 ) (constant term).

Then, apply the quadratic formula:

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

Substitute the values of ( a ), ( b ), and ( c ) into the formula:

[ x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4 \cdot 1 \cdot 6}}}}{{2 \cdot 1}} ]

[ x = \frac{{5 \pm \sqrt{{25 - 24}}}}{2} ]

[ x = \frac{{5 \pm \sqrt{1}}}{2} ]

[ x = \frac{{5 \pm 1}}{2} ]

This gives two roots:

[ x_1 = \frac{{5 + 1}}{2} = 3 ]

[ x_2 = \frac{{5 - 1}}{2} = 2 ]

So, the roots of ( y = x^2 - 5x + 6 ) are ( x = 3 ) and ( x = 2 ), both real.

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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.