Question

3rd to last :)

Answer 1

\[\sqrt{3x+7} + \sqrt{2x+6} = 4\]

Answer 2

so we move one radical to the other side: \[\sqrt{3x+7} = 4-\sqrt{2x+6}\]

Answer 3

okay good move !

Answer 4

Well move @ganeshie8
Ha ha ha..

Answer 5

square both sides: \[3x+7 = 16-8\sqrt{2x+6} +2x+6\]

Answer 6

is that right?

Answer 7

Yep..

Answer 8

thats right! just remember your conditions each step

Answer 9

Solve the right hand side part..

Answer 10

2x+6 >=0
3x+7 >=0

next....

Answer 11

I mean to say that leave square root term on one side and other terms on other side..

Answer 12

okay then we will leave the radical alone:
\[3x-2x+7-6-16 = -8\sqrt{2x+6}\]

Answer 13

Go ahead..

Answer 14

\[x-15=-8\sqrt{2x+6}\]
square both sides: \[x ^{2}-30x+225= 64+3x+7\]

Answer 15

right?

Answer 16

Something went wrong..

Answer 17

where? with the square root?

Answer 18

Right hand side..

Check it once again..

Answer 19

\[(-a \sqrt{b})^2 = a^2 b\]

Answer 20

-8square root 3x+7 = 64+3x+7

Answer 21

ah 2x+6 D:

Answer 22

Do it once again with right values..

Answer 23

\[x ^{2}-32x+154=0 \rightarrow (x+7)(x+22)=0 x=-7,-22\]

Answer 24

No...

Answer 25

Tell me simply what you got for right hand side??

Answer 26

\[x ^{2}-30x+225=64+2x+6\]

Answer 27

You added them but why??

Answer 28

See I show you more clearly..

Answer 29

ah multiply!

Answer 30

just like \[(3\sqrt{2})^{2 }= 9x2\]

Answer 31

9 x 2

Answer 32

\[\large (a \sqrt{bx + c})^2 \implies a^2 \times (\sqrt{bx+c})^2 \implies a^2 \color{green}{\times}(bx+c)\]

Answer 33

x^2-30x+225=128x+384

Answer 34

Yeah just like that..

Answer 35

This looks fine to me..

Answer 36

x^2-158x-159=0

Answer 37

Yes..

Answer 38

(x-159)(x+1) = 0 x=159, -1

Answer 39

no

Answer 40

Yes well done..

Answer 41

yes sorry lool

Answer 42

thank you so much! i think i'll go have lunch then come back in 15mins

Answer 43

See:

\[x ^{2}-32x+154=0 \rightarrow (x+7)(x+22)=0 x=-7,-22\]

Here you factorization is also wrong..

Answer 44

When you multiply you will not negative of 32 as all are positive..

Answer 45

I am just making you clear this point..
And this is also not the answer..

Answer 46

i noticed that 7+22 is not -32.. so is the whole work wrong?

Answer 47

i'll rewrite and post an attachment later.. lunch first :) thank so much btw

Answer 48

Ha ha ha..

You are now one step ahead of me..


That is really great,,'

A true leaner you are..

Answer 49

@waterineyes whole process :)
\[\sqrt{3x+7} +\sqrt{2x+6} =4 \rightarrow \sqrt{3x+7}=4-\sqrt{2x+6}\]
square both sides: \[3x+7=16-8\sqrt{2x+6}+2x+6 \rightarrow 3x-2x+7-6-16= -8\sqrt{2x+6}\]
square again: \[(x-15)^{2} = (-8\sqrt{2x+6})^{2} \rightarrow x ^{2}-30x+225= 64(2x+6)\]
solve:\[x ^{2}-158x-159=0 ----> (x-159)(x+1)=0 ---> x=159,-1\]
however when we insert the roots into the equation x=-1 only

Answer 50

Yep..
You are right..

\(x = -1\) is the only solution..