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#