Question

Algebra II help, please!

Answer 1

@johnweldon1993

Answer 2

Match the number and type of solutions for the following two functions in standard form.
f(x) = x2 + 6x – 16
g(x) = x2 +6x + 1
Then match the following descriptions of the solutions to each of the functions above.
Hint: they each have their own match.

Descriptions:
Two real irrationals solutions
Two real rationals solutions

Answer 3

why is there an x in front !! of the 2's

Answer 4

Oh sorry, that's x^2

Answer 5

do you know the quadratic equation?

Answer 6

yeh gurl you must use quadratic equation

Answer 7

ax^2+bx+c=0?

Answer 8

http://en.wikipedia.org/wiki/Quadratic_equation

Answer 9

Well we can do a little shortcut here with the quadratic formula, mainly the discriminant

Remember this part of it?

\[\large \sqrt{b^2 - 4ac}\]

if the solution to that is a whole number...we will have 2 real rational number solutions...however if we do not get a whole number (and is not negative) we will have irrational solutions

Example with the f(x)

\[\large \sqrt{6^2 - 4(1)(-16)}\]
\[\large \sqrt{36 + 64}\]
\[\large \sqrt{100}\]

is that a whole number? or a decimal?

Answer 10

yeah the just the discriminant should determine it, you're right

Answer 11

10, a whole number!

Answer 12

Exactly, so that equation f(x) will give us 2 real rational solutions

and do the same with g(x) what do you get?

Answer 13

Uhm, I'm not sure what to plug in.

Answer 14

So we have g(x)

\[\large g(x) = x^2 + 6x + 1\]

Remember that formula you had above?

\[\large ax^2 + bx + c\]

well if you compare your equation to that form...we have

a = 1
b = 6
c = 1

So if you plug those into

\[\large \sqrt{b^2 - 4ac}\]

Answer 15

√(6^2)-4(1)(1)

Answer 16

correct....which simplifies to

\[\large \sqrt{36 - 4}\]

which is \[\large \sqrt{32}\]

and is that a whole number? or a decimal?

Answer 17

5.65, so a decimal.

Answer 18

Indeed^ also known as giving us 2 irrational real number solutions :)

Answer 19

Fantastic, thank you once again!!

Answer 20

Anytime :)