How many zeroes does the function have $.25x^2-7x-12=y-4$ ?

Question

1367 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-07T00:36:13

All real quadratic functions have at most two zeroes.

For this one, if you multiply by whole thing by $4$ and set $y = 0$ :

$x^2 - 28x - 48 = -16$

$0 = x^2 - 28x - 32$

$x = (-(-28) pm sqrt((-28)^2 - 4(1)(-32)))/(2(1))$

$= (28 pm sqrt(784 + 128))/2$

$= (28 pm sqrt(912))/2 = (28 pm sqrt(57*4*4))/2$

$= (28 pm 4sqrt57)/2$

$= 14 pm 2sqrt57$

Thus the two roots are:

$color(blue)(x = 14 + 2sqrt57)$
$color(blue)(x = 14 - 2sqrt57)$

How many zeroes does the function have .25x^2-7x-12=y-4.25x^2-7x-12=y-4?