Determine the number and kind of solutions to the equation...

Question

Determine the number and kind of solutions to the equation x² - 256 = 0.

a.

2 irrational solutions

b.

2 complex solutions containing i

c.

2 rational solutions

d.

1 rational solution

Answer 1

This is all going to be based off of the discrimant b²-4ac, which can be found under the square root symbol in your quadratic formula.

If b²-4ac >0, then you will have two real and rational solutions.

If b²-4ac<0, then your square root will be a negative value, causing two complex solutions.

If b²-4ac=0, then your quadratic formula will show plus or minus 0, which causes just one real and rational solution.

The values of a, b and c come from the standard form of the equation ax²+bx+c. In this case, a = 1 and c - -256. You don't have a b value, so b = 0.

Plugging in those values to b²-4ac, (0)²-4(1)(-256) gives 1,024 which is greater than 0. So this function has two real and rational solutions (c).

If you're allowed to use a calculator, visually you could graph the function and see if has x-intercepts (solutions) at two points.. therefore has two rational solutions.

Answer 2

x² - 256 = 0

x² - 256 is a difference of squares, x² - 16², and factors into (x+16)(x-16). Hence we have:

(x+16)(x-16) = 0

The solutions are x = 16 and x = -16. Looks like 2 solutions. What kind of numbers are they, rational, irrational, or complex?