How do you find the zeros, real and imaginary, of $y=-4x^2-2x+33$ using the quadratic formula?

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Answer 1 · 2017-04-16T00:33:53

$x~~-3.133$ and $~~2.633$

The quadratic formula is bases on $ax^2+bx+c$ :
$(-b+-sqrt(b^2-(4*a*c)))/(2*a)$

So, our equation of $y=-4x^2-2x+33$ tells us $a=-4$ , $b=-2$ , $c=33$

Thus: $(-(-2)+-sqrt((-2)^2-(4*-4*33)))/(2*-4)$

$(2+-sqrt(4-(4*-4*33)))/(-8)$

$(2+-sqrt(4--528))/-8$

$(2+-sqrt532)/(-8)$

$(2+-2sqrt133)/(-8)$

$(2(1+-sqrt133))/(-8)$

$(1+-sqrt133)/-4$

That's the exact form, but the estimated form is $x~~-3.133$ and $~~2.633$

How do you find the zeros, real and imaginary, of y=-4x^2-2x+33y=-4x^2-2x+33 using the quadratic formula?