How do you find the roots, real and imaginary, of #y= 4x^2-36 # using the quadratic formula?

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Answer 1 · 2015-12-17T03:03:56

#x=+-3#

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

for the quadratic equation in the general form

#y=ax^2+bx+c#

Your equation does not seem to fit this mold, but it can be rewritten as

#y=4x^2+0x-36#

So,

#a=4#
#b=0#
#c=-36#

Plug these into the formula.

#x=(-0+-sqrt(0^2-4(4)(-36)))/(2(4))#

#x=(+-sqrt(576))/8#

#x=(+-24)/8#

#x=+-3#

These are the two real roots.