Question

Evaluate
\[\huge \left| \sqrt{7-4\sqrt{3}} \right|\]

Answer 1

that is interesting :)
show me how you would approach the problem

Answer 2

My brain thought to make that into a perfect square somehow

Answer 3

okay then try that and see

Answer 4

how to do that , is there a method

Answer 5

well you said perfect square.. what is a perfect square ?

Answer 6

how to make
\[\huge 7-4\sqrt{3}\]
into a prfect square

Answer 7

let us try a simpler problem then I will answer your question
\[\sqrt{\sqrt{16}}\]
solve that

Answer 8

that's 4th root of 16

Answer 9

2

Answer 10

16^{1/4}

Answer 11

=2

Answer 12

great!

Answer 13

@nincompoop

Answer 14

@zepdrix

Answer 15

well, even so, sometimes we need to keep going back to simpler ideas
simply saying that you are in college doesn't tell me anything how deep your math knowledge is

7-4 sqrt(3) = 4-4 sqrt(3)+3 = 4-4 sqrt(3)+(sqrt(3))^2 = (2-sqrt(3))^2

Answer 16

that was a smart move , thank you !

Answer 17

\[7-4 \sqrt{3} = 4-4 \sqrt{3}+3 = 4-4 \sqrt{3}+{\sqrt{3}}^2 = (2-\sqrt{3})^2\]

Answer 18

that was just trial and guess right

Answer 19

you split seven into 4 and 3

Answer 20

yeah we use that type of method in earlier days of factoring :P

Answer 21

HS algebra

Answer 22

ok

Answer 23

can you try the rest?

Answer 24

the problem is over

Answer 25

i mean finished

Answer 26

not quite, we still have the absolute value, and we also have the higher square root

Answer 27

2-root 3

Answer 28

okay!

Answer 29

isn't it redundant to put absolute bars as the square root already implies that you need to take positive value ?

Answer 30

yeah, it is

Answer 31

well, yeah , but that's what is the questioon

Answer 32

\[\large \sqrt{(x-3)^2} = \sqrt{(3-x)^2} = |x-3| = |3-x|\]

Answer 33

yes

Answer 34

yeah but this notion wasn't even introduced to me when I was taking algebra in HS. I learned this from reading Spivak's Calculus

Answer 35

that is a fundamental idea