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 $x = 2/6=1/3$

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