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

$x_{1,2} = \frac{-(-3)\pm\sqrt{5}}{2*1} = \frac{3pmsqrt(5)}{2}$

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