What is √1323+√588+√243?

Question

What is  √1323+√588+√243 ?

A. 27√6
B. 54√6
C. 22√3
D. 44√3
E. 54√3

BrentGMATPrepNow · Accepted Answer

ALWAYS check the answer choices first.

The answer choices tell us that the 3 radicals (square root expressions) can probably all be rewritten as EITHER 3 times some integer OR 6 times some integer. 
Since 1323 is ODD, we know that we can't rewrite it as 6 times some integer.
So, let's rewrite each expression as 3 time some integer.

√1323 + √588 + √243 = √(441 x 3) + √(196 x 3) + √(81 x 3)

Now we'll apply a nice rule that says: √(ab) = (√a)(√b)

= (√441)(√3) + (√196)(√3) + (√81)(√3)

√441 is a tough one to evaluate. However, if we recognize that √400 = 20, then perhaps √441 = 21. A quick check (21² = 441) confirms this. 
√196 is a little easier. we might already know that √225 = 15, so we might check whether √196 = 14 A quick check (14² = 196) confirms this. 
Finally, we should know that √81 = 9

So, we get....

= 21√3 + 14√3 + 9√3
= 44√3

Answer: D

shashankism · Answer

√1323+√588+√243 
= √(3^2*7^2*3)+√(2^2*7^2*3)+√(3^5) 
= 21√3+14√3+9√3 
= 44√3

VyshakhR1995 · Answer

The question can be rewritten as

21\sqrt{3} + 14\sqrt{3} + 9\sqrt{3} 
Adding we get option D as the answer

guptarahul · Answer

D √1323+√588+√243 7*3\sqrt{3} + 2*7\sqrt{3} + 9\sqrt{3} :wink:

=44\sqrt{3}

JeffTargetTestPrep · Answer

Let’s break down each root.

√1323 = √9 x √147 = 3 x √49 x √3 = 3 x 7 x √3 = 21√3

√588 = √49 x √12 = 7 x √4 x √3 = 7 x 2 x √3 = 14√3

√243 = √81 x √3 = 9√3

Thus, √1323 + √588 + √243 = 21√3 + 14√3 + 9√3.= 44√3.

Answer: D

sashiim20 · Answer

What is √1323+√588+√243?

Prime factorization of all terms
\(\sqrt{1323}\) = \(\sqrt{3^3 * 7^2}\) = 3*7 \(\sqrt{3}\) = 21\(\sqrt{3}\)
\(\sqrt{588}\) = \(\sqrt{2^2 * 3 * 7^2}\) = 2*7 \(\sqrt{3}\) = 14\(\sqrt{3}\)
\(\sqrt{243}\) = \(\sqrt{3^5}\) = 3*3 \(\sqrt{3}\) = 9\(\sqrt{3}\)

Required sum = 21\(\sqrt{3}\) + 14\(\sqrt{3}\) + 9\(\sqrt{3}\) = 44\(\sqrt{3}\)
..... Answer D ..........

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Abhishek009 · Answer

\sqrt{1323}+ \sqrt{588}+ \sqrt{243}

= \sqrt{3^3*7^2}+ \sqrt{2^2*3*7^2}+ \sqrt{3^5}

= 3*7\sqrt{3}+ 2*7\sqrt{3}+ 3^2\sqrt{3}

= 21\sqrt{3}+ 14\sqrt{3}+ 9\sqrt{3}

= 44\sqrt{3}

Thus, the answer must be (D)  44\sqrt{3}

arbazfatmi1994 · Answer

Although it can be simply solved by factorization, you should be intuitive enough to realize that the root of 3 is there in the answer choices. Hence, start to factorize by 3 first.

nammolestias · Answer

=  (√441)(√3) + (√196)(√3) + (√81)(√3) 
 = 21√3 + 14√3 + 9√3 
 = 44√3

What is √1323+√588+√243?

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What is √1323+√588+√243?

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