How do you find the zeros of $y = 3x^2 - 13x +15$ using the quadratic formula?

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Answer 1 · 2017-08-15T18:23:28

This sucker has complex roots!

$x = (-B +- sqrt((B^2-4AC)))/(2A)$

Plugging in A = 3, B = -13, C = 15 gives you:

$(13 +- sqrt(169 - 180))/6$

or $(13 +- sqrt(-11))/6$

Which is your problem, since there is no square root of a negative number.

Best you can do is rewrite it as a COMPLEX number of form $a + bi$ where i is taken to be $sqrt(-1)$ .

So you can do a little rewrite of $sqrt(-11)$ as $sqrt(-1 * 11)$ = $sqrt(11) * i$

And then evaluate it once for the positive case of the $+-$ , and once for the negative. Which gives roots of:

$13/6 + (sqrt(11)/6 * i)$ = 2.1667 + 0.553i

And

$13/6 - (sqrt(11)/6 * i)$ = 2.1667 - 0.553i

(Note that I've rounded off.

How do you find the zeros of y = 3x^2 - 13x +15 y = 3x^2 - 13x +15 using the quadratic formula?