Question

help with square root problems???

Answer 1

\[\sqrt{9y-196}+ \sqrt{196}=\sqrt{49}\]

Answer 2

one may note that \(\bf 14^2 = 196\\ 7^2 = 49\)

Answer 3

i know, so far i have \[\sqrt{9y-196} =-7\] i just dont know what to do from here

Answer 4

raise both sides to the 2nd power

Answer 5

\(\bf (\sqrt{9y-196})^2 =(-7)^2\)

Answer 6

9y-196=49?

Answer 7

yes

Answer 8

so 245/9 is the answer?

Answer 9

yes

Answer 10

i just checked it and its not right

Answer 11

hmm is this correct =>\(\bf \sqrt{9y-196}+ \sqrt{196}=\sqrt{49}\ \ \ ?\)

Answer 12

yes

Answer 13

\[\sqrt{9y+196}+14=7 \implies \sqrt{9y+196}=-7\]
square both sides

Answer 14

i figured it out the answer is 49/3

Answer 15

i mean \[\sqrt{9y-196}=-7\]
\[(\sqrt{9y-196})^2=(-7)^2\]

Answer 16

but maybe you could help with this one?\[\sqrt{x}(\sqrt{9}-\frac{ 4 }{ 3}=\sqrt{25}\]

Answer 17

theres supposed to be a closed parenthesis after 4/3

Answer 18

\(\bf \sqrt{x}\left(\sqrt{9}-\cfrac{ 4 }{ 3}\right)=\sqrt{25}\)

Answer 19

yes

Answer 20

\[\sqrt{x}(\sqrt{9}-\frac{4}{3})=\sqrt{25}\]
\[\text{since }\sqrt{9}=3,\sqrt{25}=5\]
\[\sqrt{x}(3-\frac{4}{3})=5\]
\[\sqrt{x}(\frac{5}{3})=5\]
you can take it from here

Answer 21

\(\bf \sqrt{x}\left(\sqrt{9}-\cfrac{ 4 }{ 3}\right)=\sqrt{25} \implies
\sqrt{x} = \cfrac{\sqrt{25}}{\left(\sqrt{9}-\cfrac{ 4 }{ 3}\right)}
\implies \sqrt{x} = \cfrac{\sqrt{25}}{\left(3-\cfrac{ 4 }{ 3}\right)}\)

Answer 22

thank you so much!

Answer 23

I still can't see the 49/3 as the solution to the first one!

Answer 24

nor do I

Answer 25

unless there's a typo

Answer 26

\[\sqrt{x}=5\frac{3}{5}\]
\[\sqrt x=3,x=9\]

Answer 27

its a little complicated since \[\sqrt{9y-196}=-7\] you have to square both sides,\[-(7^2)=-49\] so, 9y-196=-49
9y=147
y=49/3

Answer 28

lets take it from here
\[\sqrt{9y-196}=-7\]
\[9y-196=49\]
\[9y=245,y=\frac{245}{9}\]

Answer 29

i checked it and it works out 245/9 doesnt

Answer 30

no when you square a negetive you always get positive
\[(-a)^2=a^2\]

Answer 31

so \[(-7)^2=7^2=49\]

Answer 32

unless its-(7^2), i always learned that way too, but i mean if 49/3 works im just going to keep it. im guessing they wanted to keep the negative

Answer 33

but if you go back to you equation and plug y-49/3
theres a negetive value in the square root!!!!!!!!!!!!!!!!!1problem

Answer 34

if i get it worng then oh well, its just a homework question

Answer 35

can you just recheck the way youtyped it in correctly

Answer 36

i did it like 5 times haha im okay with it if i get one question wrong ha

Answer 37

but would you want to help me with one more? you dont have to, if you dont want to

Answer 38

\[\sqrt{135b^2c^3d}\times \sqrt{5b^2d}\]
all the exponents are confusing me

Answer 39

write evrything under the root seperatly
ie
\[\sqrt{abcd}=\sqrt{a}\sqrt{b}\sqrt{c}\sqrt{d}\]

Answer 40

okay, before multiplying the two?

Answer 41

\[\sqrt{135}\sqrt{b^2}\sqrt{c^3}\sqrt{d}\sqrt{5}\sqrt{b^2}\sqrt{d}\]

Answer 42

yes now remember
\[\sqrt{b^2}=b\]
use this

Answer 43

right do what do i do about the c^3

Answer 44

right so**

Answer 45

leave alone
but multiply like terms it
\[\sqrt{d}\times\sqrt{d}=d\]
\[\sqrt{135}\times\sqrt{5}=\sqrt{135\times 5}=\sqrt{675}\]

Answer 46

so i have \[\sqrt{675} \times b^2 \times \sqrt{c^3} \times d\]

Answer 47

is that the answer?

Answer 48

yes your a star

Answer 49

thank you!!

Answer 50

yw