Question

\[\sqrt{x} + \sqrt{x + 16 } = 3 \] find the set of solutions

Answer 1

\[\sqrt{a}+\sqrt{b}=c\]\[\implies a+b+2\sqrt{ab}=c^2.\]\[\implies c^2-a-b=2\sqrt{ab}.\]Square again: -

Answer 2

a = x
b = x+16
c = 3.

Answer 3

you'll get complex roots

Answer 4

i did that way but i was unable

Answer 5

dont forgot consider resrtictions ... easily u can see \(x\ge 0\) so \(\sqrt{x}+\sqrt{x+16} \ge 4\) and no real roots

Answer 6

@eliassaab X is not equal to 1/4 it doesn't satisfy the equation

Answer 7

@mukushla yes u r correct there is no solution.... but can u expl ur method

Answer 8

roots are complex and it is -16+\[(-16+\pm i 11.66)/ 4\]

Answer 9

Sorry I read 16 as 6

Answer 10

There is no solutions.

Answer 11

no solution
complex roots will be rejected as x must be greater or equal to 0

Answer 12

@ghazi :)

Answer 13

sqrt function has domain
\[x \ge 0.\]

Answer 14

he hasn't mentioned the condition..that's why i found the complex roots

Answer 15

got it...00

Answer 16

it does nt depends on mentioning
it is a rule for defining functions too:)
if u solve an equation:)

Answer 17

if u solve an equation
put the value in original equation & see if the function of equation is defined or nt @ghazi :D

Answer 18

hmm ..agreed .and substituting value makes no sense..i forgot the domain thing..sorry

Answer 19

no worries just practise
i also have to do that:)

Answer 20

what about the domain of complex roots?

Answer 21

complex roots aren't functions
then how can they hve dommain???????

Answer 22

i mean complex functions? when you find it's roots then how you define that function?

Answer 23

sorry i don't have enough knowledge about that
but u can take knowledge @ google.com
I mean u can search it there:)

Answer 24

thank you..i had to ask so thought to post it here...and obviously google is a universal solution for me

Answer 25

yeah:)
Best of luck :D