Question

Let ` r ` and ` s ` be the roots of \(3x^2 + 4x + 12 = 0\) Find \(r^2 + s^2\)

Answer 1

i know i have to do the viesta laws so sum of r and s is -4/3 and the product is 4

Answer 2

What are those dollar signs-

Answer 3

it was for latex but i didn't work

Answer 4

Can you retype the question

Answer 5

k
Let r and s be the roots of 3x^2 + 4x + 12 = 0. Find r^2 + s^2.

Answer 6

Can you find the roots of
\[3x^2 +4x+12\]

For example, since you can't do grouping or factoring in this, can you use the quadratic formula to find the roots?

Answer 7

yee im having trouble with that so idk :( like i can find the sum and product but thats about it

Answer 8

Do you know what the quadratic formula is

Answer 9

yes its ax^2+ax+bx+c

Answer 10

No-
\[ax^2 + bx+c\] is standard form for a quadratic equation

But the quadratic formula is
\[\frac{ (-b \pm \sqrt(b^2 -4ac)) }{ 2a }\]

Answer 11

oh cool what can i use that for?

Answer 12

ye

so, we can plug in the a, b, and c
a=3
b=4
c=12

Can you plug in these values into that

Answer 13

i need to go-

Answer 14

kek
so you would do
\[\frac{ -4\pm \sqrt(16-4(3)(12)) }{ 6 }\]
which condenses to
\[\frac{ -4\pm \sqrt(-128) }{ 6 }\]
can you simplify that?._.

Answer 15

im sorry i don't understand. i thout square roots couldnt be negative?

Answer 16

They can't, that's why they turn into i
i is \[\sqrt-1\]

Answer 17

im sorry we didn't learn this in my class ;-;

Answer 18

well, just disregard the negative for now, how would you simplify it

Answer 19

you don't need to worry about the negative in the square root because the r^2 and s^2 you are looking for are going to get rid of the square roots later anyways

Answer 20

so -4*sqrt(128 is the same as -4*8sqrt(2) which is -32sqrt(2) ?

Answer 21

wait, why multiply the -4?

Answer 22

example:
\(2 \pm3 = 5 ~or~-1\)

Answer 23

of so its adding my bad

Answer 24

It's adding adn subtracting
And then it's a _ or _ as answer

Answer 25

Just disregard that symbol for now too, xd

Let's just first simplify the sqrt(128)

Answer 26

its 8sqrt(2)

Answer 27

Yea
So, we then get
\[\frac{ -4\pm8\sqrt-2 }{ 6 }\]
Can you now simplify the, 4, 8, 6 stuff

Answer 28

Just divide the numerator and denominator by 2

Answer 29

-2+/-4sqrt(-2)/3

Answer 30

yus

this means that
\(\large\frac{-2+4\sqrt{-2}}{3}\) is one "root"
and the other root is
\(\large\frac{-2-4\sqrt{-2}}{3}\)

Answer 31

oh how would i apply r^2+s^2 to that tho

Answer 32

like i guess i plug it in but im a bit confused on how to calculate

Answer 33

\(\huge(\frac{-2+4\sqrt{-2}}{3})^2+(\frac{-2-4\sqrt{-2}}{3})^2\)

Answer 34

i'm sorry i don't understand on how to calculate that :(

Answer 35

\(\color{#0cbb34}{\text{Originally Posted by}}\) Angle
\(\huge(\frac{-2+4\sqrt{-2}}{3})^2+(\frac{-2-4\sqrt{-2}}{3})^2\)
\(\color{#0cbb34}{\text{End of Quote}}\)

\(\large(\frac{-2+4\sqrt{-2}}{3} \times \frac{-2+4\sqrt{-2}}{3})+(\frac{-2-4\sqrt{-2}}{3}\times \frac{-2-4\sqrt{-2}}{3})\)

Answer 36

ok so would it be (-2+8)/3^2=6/3^2=2^2=4 and also (-2-8/3)^2=-10/3=-100/9 did i do something wrong???

Answer 37

\((-2+4\sqrt{-2})(-2+4\sqrt{-2}) = 4 + 8\sqrt{-2}+8\sqrt{-2}+16(-2)\)

Answer 38

-32+4+16sqrt(-2)=-28+16sqrt(-2)

Answer 39

yes
and do the same with the other set

Answer 40

i got 4-8sqrt(-2)-8sqrt(-2)+8=4-8-16sqrt(2)=-4-16sqrt(-2)

Answer 41

i may have made a typo

Answer 42

is it 4+8sqrt(-2)+8sqrt(-2)+8 and 12+8sqrt(-2)

Answer 43

4-8sqrt(-2)-8sqrt(-2)+16(-2) = 4 - 16sqrt(-2) - 32

Answer 44

then i add those togheter 4 - 16sqrt(-2) - 32+-28+16sqrt(-2)=-56

Answer 45

and so uts -56/3? or -18 2/3

Answer 46

over 9?

Answer 47

yep i got it write thx so much ^-^

Answer 48

:)