How do you find the zeros, real and imaginary, of $y=3x^2+17x-3$ using the quadratic formula?

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Answer 1 · 2016-03-12T17:05:01

The quadratic has 2 real roots at $x_+=(5sqrt(13)-17)/6~=0.171$ and $x_(-)=(-17-5sqrt(13))/6~=-5.84$

To begin, we must put our quadratic in the standard form (which is it already) and set $y$ equal to zero.

$3x^2+17x-3=0$

The quadratic formula uses the form:

$ax^2 +bx+c=0$

where a,b, and c can be gotten from the equation above. The roots are given by:

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

If the roots are complex, we will get a negative number under the square root. In our case the roots are given by:

$x=(-17+- sqrt(17^2 -4*3*(-3)))/ (2*3)$

$x = (-17+- 5sqrt(13))/ (6)$

$x_+=(5sqrt(13)-17)/6~=0.171$

$x_(-)=(-17-5sqrt(13))/6~=-5.84$

How do you find the zeros, real and imaginary, of y=3x^2+17x-3y=3x^2+17x-3 using the quadratic formula?