How do you find the roots, real and imaginary, of #y= 5x^2 - 13x + 4+x(x-1) # using the quadratic formula?

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Answer 1 · 2017-08-14T17:59:43

See below.

Solve

#5x^2-13x+4+x(x-1) = 6x^2-14x+4 = 2(3x^2 -7x+2) = 0#

Which is equivalent to

#3x^2-7x+2 =0#

We were told to use the quadratic formula, so

#x = (-(-7)+-sqrt((-7)^2-4(3)(2)))/(2(3))#

# = (7+-sqrt25)/6 = (7+-5)/6#

#x=12/6=2# or #x = 2/6=1/3#