How do you find the roots, real and imaginary, of #y=-14 x^2 +18x +16-(x-3)^2 # using the quadratic formula?

1 Answer
sente
Aug 6, 2016

#x in {4/5-sqrt(249)/15, 4/5+sqrt(249)/15}#

Explanation:

The quadratic formula gives us the solutions to a quadratic equation of the form #ax^2+bx+c=0#. Thus, to use it, we must work the equation into that form.

#-14x^2+18x+16-(x-3)^2#

#= -14x^2+18x+16-(x^2-6x+9)#

#= -14x^2+18x+16-x^2+6x-9#

#=-15x^2+24x+7#

Thus, we can now look for the roots as the solutions to #ax^2+bx+c=0#, with #a=-15, b=24, c=7#

Plugging our values into the formula, we get

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

#=(-24+-sqrt(24^2-4(-15)(7)))/(2(-15))#

#=-(-24+-sqrt(576+420))/30#

#=(24+-sqrt(996))/30#

#=4/5+-sqrt(249)/15#

Thus, the roots of the given equation are #{4/5-sqrt(249)/15, 4/5+sqrt(249)/15}#

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