Question

(sqrt2z+6)+(sqrtz+4)-(sqrt8z+9)=0

Answer 1

\((\sqrt{2z}+6)+(\sqrt{z}+4)-(\sqrt{8z}+9)=0\)
Is this the question?

Answer 2

yes thats the question to be solved. thnx

Answer 3

\(\sqrt{8z}=\sqrt{2*2*2*z}\)
\(=2\sqrt{2z}\)
\((\sqrt{2z}+6)+(\sqrt{z}+4)-(2\sqrt{2z}+9)=0\)
\(\sqrt{2z}+6+\sqrt{z}+4-2\sqrt{2z}-9=0\)
This simplifies to
\(1+\sqrt{z}-\sqrt{2z}=0\)
\(\sqrt{2z}-\sqrt{z}=1\)
Factor out \(\sqrt{z}\). Then u will get
\(\sqrt{z}(\sqrt{2}-1)=1\)
Square both sides and solve for z.
Can u do it @yayimoakamie?

Answer 4

ok thnx let me give it a try

Answer 5

sure:)

Answer 6

is the final answer z=1

Answer 7

Hmm no.

Answer 8

Did u round off ur answr?

Answer 9

@yayimoakamie

Answer 10

no

Answer 11

i got it wrong then

Answer 12

yup.
\((\sqrt2-1)^2=(\sqrt2)^2-2*\sqrt2*1+1^2\)
\(=2-2\sqrt2+1\)
\(=3-2\sqrt2\)
\((\sqrt z)^2=z\)
\(z(3-2\sqrt2)=1\)
\(z=\LARGE\frac{1}{3-2\sqrt2}\)
Getting this @yayimoakamie?

Answer 13

Thnx Ajprincess for your help appreciated

Answer 14

yw:) do u understand it?

Answer 15

honestly i am lost

Answer 16

Tell me where u r lost?

Answer 17

why is the square root not affecting the constants but just the variables

Answer 18

Sorry. I dnt get ur question.

Answer 19

can u plz show where it is nt being affected?

Answer 20

@yayimoakamie

Answer 21

sorry what i mean is d constants 6+4+9 were not affected by the root

Answer 22

in ur original question square root is for only z right? That is y they are not being affected.

Answer 23

when u solve an equation u solve for the variable. the variable should be alone without square root or anything.