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

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Answer 1 · 2018-05-06T13:07:53

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:

If #Delta>0#, then the square root is well defined and not null, so you have two solutions.
If #Delta=0#, the square root is zero as well. So, adding or subtracting it makes no difference, and you have two coincident solutions.
If #Delta < 0#, the square root is not defined, and you have no solutions (at least using real numbers).

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}#