2 questions: find the value of Ax + Bx^2 - C; solve for x: 3x^2 - 14x -5 <0; help please! :)?

Question

2 questions

1) find the value of:
Ax + Bx^2 - C
if
3x=2
C=4
A-B=C
A+B=2C

answer: 8/9

--------------

2) solve for x:
3x^2 - 14x -5 <0

answer:
-1/3 < x < 5

----------

how do you get those answers? please explain step by step. thanks :)

Answer 1

Ok, quesiton 1.

A - B = C
A + B = 2C

C = 4 so

1) A - B = 4
2) A + B = 2(4) = 8

A = B + 4 (see 1 above)

2) becomes (plug back in with only one variable)
B + 4 + B = 8
2B + 4 =8
2B = 4
B = 2

Then go back to 1) or 2) and plug in 2 for B

1) A -2 = 4, A = 6
2) A + 2 = 8 A = 6

So the first equation now becomes

6x +2x^2 - 4

and you know that 3x = 2 so x = 2/3

6 (2/3) + 2(2/3)^2 - 4

now it is math

Ax ->6 (2/3) = 12/3 = 36/9

Bx^2->2(2/3)^2 = 2(4/9) = 8/9

C->4 = 36/9

36/9 + 8/9 - 36/9 = your answer.

Forget about that lady.


Question 2.

I hope you know the FOIL (first, outside, inside, last) method for factoring quadratic equations. sometimes you have to play the guess and check game to figure out what the anwsers are.

in this case

3x^2 -14x - 5 < 0

(3x [ ] ) * (x [ ] ) you know that you have to muliply x and 3x.

Foil first

3x * x = 3x^2

now you need to guess and check to see. you have a -14x so obvious the 3*5 will have a minus.

fOil outside

(3x [ ] ) * (x [ - ] 5 )

and the x to add to -15x

foIl inside

(3x [ +] 1) * (x [ - ] 5 )

and

foiL last to check

-5 * 1 = -5

so you have your two sets now

(3x + 1) (x - 5) < 0


so

(x - 5) < 0

x < 5

(3x + 1) < 0 remember that when you multiply by a negative

x < -1/3

Answer 2

Please this is not a place to answer your homework or to help you,it's homework and you should solve it yourself or ask an adult gosh.