Question

I need help with algebra again

Answer 1

\[-\sqrt{6x+1}-5=-7-\sqrt{3x+1}\]

Answer 2

thinking

Answer 3

lol ok

Answer 4

ok get the numbers to the right and those square roots to the left

Answer 5

hint:
we can rewrite your equation as below:
\[\Large 7 - 5 = \sqrt {6x + 1} - \sqrt {3x + 1} \]

Answer 6

half of me is like square both sides to get rid of those square roots. after that step

Answer 7

after a first square, we get:
\[\Large 4 = 6x + 1 + 3x + 1 - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \]
please simplify @MathHater82

Answer 8

oh golly .

Answer 9

o.o im lost

Answer 10

hint:
here are the next steps:
\[\Large \begin{gathered}
4 = 9x + 2 - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \hfill \\
\hfill \\
2 = 9x - 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} \hfill \\
\end{gathered} \]

Answer 11

ok but how did you do the step before that

Answer 12

\[\Large 2= \sqrt {6x + 1} - \sqrt {3x + 1} \]
\[\Large 2^2= (\sqrt {6x + 1} - \sqrt {3x + 1})^2 \]

\[\Large 4= (\sqrt {6x + 1} - \sqrt {3x + 1})(\sqrt {6x + 1} - \sqrt {3x + 1}) \]
and then expand ?

Answer 13

with foil?

Answer 14

since I have computed the square of both sides, like this:
\[\Large {\left( {7 - 5} \right)^2} = {\left( {\sqrt {6x + 1} - \sqrt {3x + 1} } \right)^2}\]

Answer 15

oh ok

Answer 16

good luck to that on the right hand side that's killer

Answer 17

next, we can write:
\[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2\]

Answer 18

and, taking the square of both sides again, we have:
\[\Large {\left( {2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} } \right)^2} = {\left( {9x - 2} \right)^2}\]

Answer 19

or, after a simplification:
\[\Large 4\left( {18{x^2} + 9x + 1} \right) = 81{x^2} + 4 - 36x\]

Answer 20

I have applied the foil method inside the radical at left side

Answer 21

so, please continue @MathHater82

Answer 22

oh yes it's getting to be a bit easier. . . . after those square roots are gone.

Answer 23

ok im just trying to follow this all on my white board i have here

Answer 24

give me a sec

Answer 25

\[36x+4=9x^2-36x+4\] \[72x=9x^2\]

Answer 26

ok! correct!
now try to solve that equation

Answer 27

x=8

Answer 28

hint:
we have the subsequent steps:
\[\Large \begin{gathered}
9{x^2} - 72x = 0 \hfill \\
\hfill \\
9x\left( {x - 8} \right) = 0 \hfill \\
\end{gathered} \]

Answer 29

maybe we can shift the 9x^2 and se the equation equal to 0. whoops backwards

0 = 9x^2-72x

Answer 30

yes solve for x after what @Michele_Laino typed
there's 2 x's to solve.. so first we have
9x=0

and then x-8=0

Answer 31

x=8 is correct!
nevertheless also x=0 is a solution
that's right! @UsukiDoll

Answer 32

x = 0,8 but uh oh only one x works

Answer 33

so, x=0 is a solution or not? @MathHater82

Answer 34

please recall this step:
\[\Large 2\sqrt {\left( {6x + 1} \right)\left( {3x + 1} \right)} = 9x - 2\]

Answer 35

I think plugging in x = 0 in the equation that we had started with is easier.

Answer 36

noew, the left side is positive, since it is the positive square root, so also the right side has to be positive. Namely we have:
\[\Large 9x - 2 > 0\]

Answer 37

now*

Answer 38

or:
\[\Large x > \frac{9}{2}\]
now x=0, doesn't check that condition, so x=0 can not be an acceptable solution

Answer 39

also when x = 0
\[-\sqrt{1}-5=-7-\sqrt{1}\]
-1-5=-7-1
\[-6 \neq -8\]

Answer 40

yes! that is another possible way to solve the question

Answer 41

mathematicians call x=0 an extraneous solution

Answer 42

I don't really understand why x=0

Answer 43

since x=0 and x=8 are the solutions of this equation:
\[\Large 9{x^2} - 72x = 0\]

Answer 44

we obtained two solutions, but 1 works.

Answer 45

how did we obtain the solution 0 and which one worked

Answer 46

the equation \[\Large 9{x^2} - 72x = 0 \] got factored.. we can factor a 9x out

Answer 47

\[\large 9x(x-8)=0\]
then we solve for each case
9x=0 and x-8=0
that leads us to x =0,8 but only 1 works.

Answer 48

ok

Answer 49

furthermore. looking at your original equation also these conditions have to be checked:
\[\Large \begin{gathered}
6x + 1 \geqslant 0 \hfill \\
3x + 1 \geqslant 0 \hfill \\
\end{gathered} \]
for existence of both radicals, which are automatically satisfied since:
\[\Large 8 > \frac{9}{2}\]

Answer 50

ok cool i get it thanks for the help