How do you find the roots, real and imaginary, of #y= 3x^2-5x- 1 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y= 3x^2-5x- 1   # using the quadratic formula?

Savannah Anderson · Accepted Answer

#x=5/6+(sqrt3)/2,# #5/6-(sqrt 3)/2#

#y=3x^2-5x-1# is a quadratic equation in standard form, #y=ax+bx+c#, where #a=3, b=-5, c=-1#, and #y=0#.

The roots are the solutions for #x#.

Quadratic Formula

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

Substitute the known values into the equation.

#x=(-(-5)+-sqrt((-5)^2-(4*3*-1)))/(2*3)#

Simplify.

#x=(5+-sqrt(25+12))/6#

Add #25# and #12#.

#x=(5+-sqrt27)/6#

Simplify #sqrt(27)#.

#sqrt27=sqrt(3*3*3)=sqrt(3^2*3)=3sqrt 3#

#x=(5+-3sqrt3)/6#

Write the value for #x# as two fractions.

#x=5/6+-(3sqrt3)/6#

Simplify #3/6# to #1/2#.

#x=5/6+-(sqrt3)/2#

Isaiah Anthony · Answer

undefined To find the roots of the quadratic function $ y = 3x^2 - 5x - 1 $ using the quadratic formula, which is given by:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Where $ a = 3 $, $ b = -5 $, and $ c = -1 $ are the coefficients of the quadratic equation.

Plugging these values into the quadratic formula:

$$ x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4 \cdot 3 \cdot (-1)}}}}{{2 \cdot 3}} $$

$$ x = \frac{{5 \pm \sqrt{{25 + 12}}}}{{6}} $$

$$ x = \frac{{5 \pm \sqrt{{37}}}}{{6}} $$

So, the roots of the quadratic function $ y = 3x^2 - 5x - 1 $ are:

$$ x = \frac{{5 + \sqrt{{37}}}}{{6}} $$ (real root)
$$ x = \frac{{5 - \sqrt{{37}}}}{{6}} $$ (real root)