How do you find the roots, real and imaginary, of $y= 4x^2-36$ using the quadratic formula?

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Answer 1 · 2015-12-17T03:03:56

$x=+-3$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

for the quadratic equation in the general form

$y=ax^2+bx+c$

Your equation does not seem to fit this mold, but it can be rewritten as

$y=4x^2+0x-36$

So,

$a=4$
$b=0$
$c=-36$

Plug these into the formula.

$x=(-0+-sqrt(0^2-4(4)(-36)))/(2(4))$

$x=(+-sqrt(576))/8$

$x=(+-24)/8$

$x=+-3$

These are the two real roots.

How do you find the roots, real and imaginary, of y= 4x^2-36 y= 4x^2-36 using the quadratic formula?