Question

Find the largest (most positive) value of t satisfying:
t2 + 16 t - 27 = 12 t - 6

Answer 1

solve for t first. subtract 12t from both sides..what do you get?

Answer 2

t^2+4t-27

Answer 3

t^2 + 4t - 27 = -6

don't forget that right side

now add 6 to both sides

Answer 4

t^2+4t-21=0

Answer 5

right. now factor out that equation

Answer 6

how do you factor... thats what all my equations come to and i just dont remember how to do it

Answer 7

if you're not used to factoring...you can just use the quadratic formula..do you know the formula?

Answer 8

yes i found it
let me try it out

Answer 9

i got 4 ?

Answer 10

no it's not 4

Answer 11

let me give you a hint

t^2 + 4t - 21 = 0

factor it out...

(t+7)(t-3) = 0

Answer 12

solve for t

Answer 13

3 and -7

Answer 14

so which one is the positive?

Answer 15

3
Awesome so for this equation 3x^2-16x+16=0 it would be 16 and -1 because 16 mult by -1 is 16 so then it would be (x-1)(x+16)

Answer 16

No, (x-1)(x+16) = x^2 +15x -16. You are looking for 3x^2-16x+16

Answer 17

don't disregard that 3 beside x^2

Answer 18

can i multiply the 16's by 3?

Answer 19

just break down the middle term

3x^2 - 12x - 4x + 16 = 0

factor that out

Answer 20

i have to go now so i hope you get this @kaylakelly13

Answer 21

thank you so much

Answer 22

welcome

Answer 23

In FOIL:

First
Outside
Inside
Last

You want the Last terms to equal 16 when multiplied.

So, maybe it is (3x + 2)(x + 8). Because 2 x 8 = 16

Answer 24

Any ideas?

Answer 25

well wouldnt at least one need to be negative ?

Answer 26

Yes.

Answer 27

But, 16 is positive. So what does that mean?

Answer 28

that they are both negative

Answer 29

Correct! So far you know this...

(3x - )(x - ), fill in the blanks.

Answer 30

I suggested 2 and 8 before:

(3x - 2)(x - 8)

Answer 31

could it be (3x-4)(x-4)?

Answer 32

Multiply it out and tell me what you get.

Answer 33

i got 3x^2-12x-4x+16 but i would get the same for (3x-2)(x-8)

Answer 34

Let's look at (3x-2)(x-8):

Use FOIL
FIRST: 3x * x = 3x^2, that works
OUTSIDE: 3x * -8 = -24x
INSIDE: -2 * x = -2x
LAST: -2 * -8 = 16

So, we get 3x^2 -24x -2x +16

Answer 35

The OUTSIDE and INSIDE are not correct. So it looks like you were right: the answer is -4 and -4.

Answer 36

but my answer has to be positive

Answer 37

What do you mean? Why does it have to be positive?

Answer 38

This is the original equation:

3x^2-16x+16=0

Answer 39

yea

Answer 40

3x^2-16x+16 = (3x^2 - 4)(x - 4) = 3x^2-12x-4x+16 = 3x^2-16x+16

Answer 41

Negative 4 times negative 4 = positive 16; is that what you mean?

Answer 42

Or do you mean that x has to be positive?

Answer 43

Find the largest (most positive) value of x satisfying

Answer 44

Oh ok

Answer 45

So, we know that 3x^2-16x+16 = (3x^2 - 4)(x - 4) right?

Answer 46

yeah

Answer 47

And (3x^2 - 4)(x - 4) = 0 in the original question

Answer 48

So basically, for what values of x does (3x^2 - 4)(x - 4) = 0?

Answer 49

Whenever you multiply a number by zero you get 0

Answer 50

So, when (3x^2 - 4)(x - 4) = 0

(3x^2 - 4) could be zero
(x - 4) could be zero

Answer 51

Set (3x^2 - 4) equal to zero and solve for x
Then set (x-4) equal to zero and solve for x

The more positive x is your answer

Answer 52

x - 4 = 0

Add 4 to both sides

x = 4

Answer 53

4 would be the answer cus the other one would be teh square root of 3/4

Answer 54

3x^2 - 4 = 0

Add 4 to both sides

3x^2 = 4

Divide by 3

x^2 = 4/3

Square root

x = SQRT(4/3), which simplifies into 2SQRT(3)/3 (DON't WORRY ABOUT THIS, IT'S LESS THAN 4)....

So, x = 4 is largest

Answer 55

thank you sooo much! this really helped me