Quadratic functions = confusing .. can you tell?

Question

Quadratic functions = confusing .. can you tell?



11. x^2 - 6x + 6 = 0

12. x^2 + 2x = 15

13. x^3 + 2x - 5 = 0

14. 2x^2 + 8x - 10 = 0

15. 4x^2 + 4x = 3


Anyone who actually can solve these problems is my official hero =). lol. Because I have a quiz soon and i'm tryin to study these problems. But it doesn't help when you don't have the right answers. haha.


Thanks a bunch guys.

Answer 1

11 isn't factorable so you must use quadratic equation.

(6 +/- sqrt(36 - 24))/2 = (6 +/- 2sqrt(3))/2 = 3 +/- sqrt(3)

12 factors to (x + 5)(x - 3)

13 (I'm assuming that should be x^2?) doesn't factor so:

(-2 +/- sqrt(4 + 20))/2 = (-2 +/- 2sqrt(6))/2 = -1 +/- sqrt(6)

14 divide by 2 first to get x^2 + 4x - 5 = 0

This factors to (x + 5)(x - 1)

15 (4x^2 + 4x - 3 = 0) use quadratic:

(-4 +/- sqrt(16 + 48))/8 = (-4 +/- 8)/8 = -12/8 or 4/8, or -3/2 and 1/2.

Answer 2

Well, you have to use the quadratic formula to solve these (or you can factor). The quadratic formula is as follows:
(-b +/- sqrt(b^2 - (4ac)) / 2a
a is the coefficient of the squared, b is of the plain x, and c is the constant (no x)
I will do number 11 as an example.
(-6 +/- sqrt( (-6^2) - (4*1 * 6)) / 2(1)
(-6 +/ - sqrt(36 - 24)) / 2
(-6 +/- sqrt(12)) / 2
(-6 + 4) /2 or (-6 - 4) / 2
-2 /2 or -10/2
= -1 or -5

Answer 3

I will give you a hand....but is up to you to understand the principle.

11. x^2 - 6x + 6 = 0

you need to use the quadratic equation,

x^2 - 6x + 9 = 3
(x-3)(x-3) = 3
(x-3)^2 = 3
(x-3) = +/-sqrt(3)

x = 3 +/- sqrt (3)

12. x^2+2x-15= 0

this one is an easy one -15 = 5*(-3)

or (x-3)(x+5)

13. x^3 + 2x - 5 ...please check for typo

14. 2x^2 + 8x - 10 = 0
Divide by 2 the entire expression,

x^2 + 4x - 5 = 0

or (x-1)(x+5) = 0

15. 4x^2 + 4x -3 = 0
Same as above, divide by 4,

x^2 + x - 3/4 = 0

complete the square,

x^2 + x + (1/2)^2 = 1.00

(x+0.5)(x+0.5) = 1.00
(x+0.5)^2 = 1.00
(x+0.5) = +/-1.00 sqrt(1) = +/-1

x = -0.5 +/-1 , or
x= -1.5 ; 0.5

Good luck.

Answer 4

There are two ways to solve a quadratic f'n:
1. Use the quadratic formula
2. Factor
Note that, with both, you must have the equation equal to zero.The second method only works with some quadratic functions, so I would suggest going straight for the first. If you don't know what the quadratic formula is, then look it up on wikipedia. I would write it out for you, but it is really complicated and I don't have the symbols to do it on this keyboard. Good luck with your test!

Answer 5

Given a chain f(n) for n = 0,...,N say n(i) is the ith value of n. calculate f(n(i+a million)) - f(n(i)) for each sequential pair. call the consequence g(m) (nonetheless a chain) and repeat the comparable exercize: g(m(j+a million)) - g(m(j)) If the ensuing series is continuous, then f(n) might properly be modeled by using a quadratic function of n. (this is a quadratic function in n). that's in easy terms a diverse thank you to instruct a). Your way is greater useful. adequate yet no longer mandatory. So, if ameliorations are consistent ( for = a million, i+a million,a million+2, i+3,...) then f = mx+b if ameliorations of ameliorations are consistent then f= ax² +bx+c if diff of diff of diff are consistent then f = ax^3+bx² + cx +d and so on., and so on.

Answer 6

Plug them into the quadratic formula:
x=(-b+-sqrt(b^2-4ac))/2a

a, b, c are the coefficients of the original function. So in number 11, a=1, b=-6, c=6.

Answer 7

11. x^2 - 6x + 6 = 0
(x-4)(x-2)=0
x=4, 2

12. x^2 + 2x = 15
x^2+2x-15=0
(x+5)(x-3)=0
x=3, -5

13. x^3 + 2x - 5 = 0
x=(-2+/-√(4+20))/2
x=-1+/-√6
x=-1-√6, -1+√6

14. 2x^2 + 8x - 10 = 0
x^2+4x-5=0
(x+5)(x-1)=0
x=1, -5

15. 4x^2 + 4x = 3
4x^2+4x-3=0
(2x-1)(2x+3)=0
x=1/2, -3/2

Answer 8

Look up the quadratic equation...it's pretty simple. just a plug in formula. I would but I don't have time.