How do you factor $4z^{2} + 33= 24z$ ?

Question

1490 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-16T02:45:34

$color(purple)(z = 3 + sqrt3/2, color(blue)(3 - sqrt3 /2$

$4z^2 + 33 = 24x$

$4z^2 - 24 z + 33 = 0$

$"It's in the form " a x^2 + b x + c = 0$

$a = 4, b = -24, c = 33$

$"roots are " z = (-b +- sqrt(b^2 - 4 a c) / (2 a))$

$"Discriminant " d = sqrt(b^2 - 4 a c)$

$d = sqrt((-24)^2 - 4 * 4 * 33)) = sqrt(48)$

Hence both the roots are real.

$:. z = (24 +- sqrt48) / 8 = 3 +- sqrt3/2$

$color(purple)(z = 3 + sqrt3/2, color(blue)(3 - sqrt3 /2$

Answer 2 · 2018-07-16T11:30:37

$(2z-6+sqrt3)*(2z-6-sqrt3)=0$

$4z^2+33=24z$

$4z^2-24z+33=0$

$4z^2-24z+36-3=0$

$(2z-6)^2-(sqrt3)^2=0$

$(2z-6+sqrt3)*(2z-6-sqrt3)=0$

How do you factor 4z^{2} + 33= 24z4z^{2} + 33= 24z?