How do you find the roots, real and imaginary, of #y= (45-x)x # using the quadratic formula?

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Answer 1 · 2016-05-07T03:23:29

#x = 0#, #x = 45#

Without using the quadratic formula, simply equate the equation to #0#

#(45 - x)x = 0#

For a product to be #0#, either factor should be 0

#=> 45 - x = 0#
#=> x = 45#

#=> x = 0#

If you really need to get the zeroes using the quadratic formula.

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

#y = (45 - x)x#

#=> y = -x^2 + 45x + 0#

#=> a = -1#
#=> b = 45#
#=> c = 0#

#x = (-45 +- sqrt(45^2 - 4(-1)(0)))(2(-1))#

#=> x = (-45 +- sqrt(45^2 - 0))/-2#

#=> x = (-45 +- 45)/-2#

#=> x = (-45 + 45)/-2#
#=> x = 0/-2 = 0#

#=> x = (-45 - 45)/-2#
#=> x = -90/-2#
#=> x = 45#