Question

square root of 64x^6

Answer 1

the square root of 64 is ______

Answer 2

8

Answer 3

the square root of x^6 is ______

Answer 4

that i do not know.

Answer 5

what you do is divide the exponent by 2

Answer 6

in this case, the exponent is 6, so 6/2 = 3

Answer 7

the square root of x^6 is x^3

Answer 8

put this all together to get

\[\large \sqrt{64x^6} = 8x^3\]

Answer 9

so whatever the x^ is you / it by 2
for the square root

Answer 10

thanks

Answer 11

exactly

Answer 12

another example

\[\large \sqrt{x^{14}} = x^{14/2} = x^7\]

\[\large \sqrt{x^{14}} = x^7\]

Answer 13

even when the ^ is a square number like 16,36 ect. and what if its a odd number

Answer 14

like ^25 would it be 12.5

Answer 15

it would be

\[\large \sqrt{x^{25}} = x^{25/2}\]

or

\[\large \sqrt{x^{25}} = x^{12.5}\]

since 25/2 = 12.5

Answer 16

alternatively, you can say this

\[\large \sqrt{x^{25}} = x^{25/2}\]

\[\large \sqrt{x^{25}} = x^{(24+1)/2}\]

\[\large \sqrt{x^{25}} = x^{24/2}*x^{1/2}\]

\[\large \sqrt{x^{25}} = x^{12}*x^{1/2}\]

\[\large \sqrt{x^{25}} = x^{12}*\sqrt{x}\]

Answer 17

ok what about (x+7)^2=-13

Answer 18

what do you get when you square any number

is that result ever negative?

Answer 19

no so is that mean there is no answer

Answer 20

there's no real numbered answer

Answer 21

when you start to study imaginary and complex numbers, it will change how you answer

but for now, there are no real solutions

Answer 22

you wouldn't turn (x+7)^2=-13 into x^2+14x+49=-13

Answer 23

you could, but that's going in the wrong direction

Answer 24

if it was a positive number would you square that number and have x+7= the square root of 13

Answer 25

yes this would work if 13 was positive, that would give you 2 real solutions

Answer 26

then subtract the 7 from the square root of 13 to figure out x

Answer 27

yeah if you had (x+7)^2 = 13

then you would get

x = -7 + sqrt(13) or x = -7 - sqrt(13)

Answer 28

so to the nearest thousandth would be 3.606
then i would -7+3.606 or -7+-3.606

Answer 29

yeah but remember this only works if the 13 is positive

in the original problem, the 13 is negative

Answer 30

ok thanks for the help

Answer 31

yw

Answer 32

if your still there how would i solve 7p^2=-9p-2
since the answer is a - dose that mean it dose not have a true number answer.
I have to use the quadratic formula

Answer 33

yeah i recommend using the quadratic formula

Answer 34

if i remember it is x=-b square root of b^2-4ac and all of that over 2a

Answer 35

-b +- sqrt(b^2 - 4ac) all over 2a

Answer 36

once i get the sort(b^2-4ac) i then *the -b first then all of that is over 2a?

Answer 37

right now i have -(-12)sqrt(144-4*3*-3 all over 6
the sqrt would be 48
so i have 12 sqrt 48 all over 6

Answer 38

you are forgetting the plus or minus

Answer 39

between -b and the square root

Answer 40

12+or-the sqrt of 48 all over 2a

Answer 41

you're solving 7p^2=-9p-2 still right? or is this a different problem?

Answer 42

sorry my bad i am working on 3x^2-12x-3=0

Answer 43

ok one sec

Answer 44

this is what you should get


\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large x = \frac{-(-12)\pm\sqrt{(-12)^2-4(3)(-3)}}{2(3)}\]


\[\Large x = \frac{12\pm\sqrt{144-(-36)}}{6}\]


\[\Large x = \frac{12\pm\sqrt{144+36}}{6}\]


\[\Large x = \frac{12\pm\sqrt{180}}{6}\]


\[\Large x = \frac{12+\sqrt{180}}{6} \ \text{or} \ x = \frac{12-\sqrt{180}}{6}\]


\[\Large x = \frac{12+6\sqrt{5}}{6} \ \text{or} \ x = \frac{12-6\sqrt{5}}{6}\]


\[\Large x = \frac{6(2+\sqrt{5})}{6} \ \text{or} \ x = \frac{6(2-\sqrt{5})}{6}\]


\[\Large x = 2+\sqrt{5} \ \text{or} \ x = 2-\sqrt{5}\]

Answer 45

how do I graph quadratic equation y=3x^2

Answer 46

by plotting a series of points

Answer 47

each point is found by plugging in values of x to get corresponding values of y

Answer 48

all i need is 2 different values for x like 1 and 2?

Answer 49

and figure what y= is x is 1 and if x is 2

Answer 50

I would do more than 2 points

Answer 51

would you do 4 points

Answer 52

4 might be good

Answer 53

when graphing y=(x-2)^2 I used the x value of 1,2,3,4 when i got the y value it was1,0,1,4
that made two parabolas one going facing up and one facing down? what dose this mean

Answer 54

you should only get one parabola

Answer 55

that is opening upward

Answer 56

i see were i messed up, I need to plot them in order value