How do you find the zeros, if any, of #y=x^2-3x+1# using the quadratic formula?

Answer 1

The solutions are #\frac{3pmsqrt(5)}{2}#

Given a quadratic equation

#ax^2+bx+c=0#

the quadratic formula states that the solutions, if any, are

#x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}#
The quantity #\Delta = b^2-4ac# is called Determinant, and you can tell the number of solutions by its sign:
In your case, #a=1#,#b=-3# and #c=1#. So, you have
#Delta = (-3)^2 - 4 * 1 * 1 = 9-4 = 5#, so you have two solutions, namely
#x_{1,2} = \frac{-(-3)\pm\sqrt{5}}{2*1} = \frac{3pmsqrt(5)}{2}#
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Answer 2

To find the zeros of ( y = x^2 - 3x + 1 ) using 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:

( a = 1 ) ( b = -3 ) ( c = 1 )

Substitute these values into the quadratic formula:

( x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4(1)(1)}}}}{{2(1)}} )

( x = \frac{{3 \pm \sqrt{{9 - 4}}}}{{2}} )

( x = \frac{{3 \pm \sqrt{5}}}{{2}} )

So the zeros are ( x = \frac{{3 + \sqrt{5}}}{{2}} ) and ( x = \frac{{3 - \sqrt{5}}}{{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.

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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