((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2

Question

(\sqrt{15 - 4\sqrt{14}} + \sqrt{15 + 4\sqrt{14}})^2= A. 28 B. 30 C. 32 D. 34 E. 36 roots.png

Bunuel · Accepted Answer

Apply  (a+b)^2=a^2+2ab+b^2 :

(\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2=

=(15+4\sqrt{14})+2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}+(15-4\sqrt{14})=

=30+2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}} .

Now, apply  (a+b)(a-b)=a^2-b^2 :

30+2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}=

=30+2\sqrt{15^2-16*14}=

=30+2*1=

=32 .

Answer: C.

Hope it's clear.

PareshGmat · Answer

I was able to solve it in 60 seconds

4\sqrt{14} = \sqrt{224}

15 = \sqrt{225}

= (\sqrt{\sqrt{225} + \sqrt{224}} + \sqrt{\sqrt{225} - \sqrt{224}})^2

= \sqrt{225} + \sqrt{224} + 2 \sqrt{(\sqrt{225} + \sqrt{224})( \sqrt{225} - \sqrt{224})} + \sqrt{225} - \sqrt{224}

= 15 + 2 + 15

= 32

Answer = E

GSBae · Answer

The trick is knowing that anytime you have a number like  200^2 - (199)(201)  the answer will always be 1.

Proof:

x^2 - (x+1)(x-1) = x^2 - (x^2 - 1) = 1.

Stiv · Answer

This answer is quite time consuming. Is there any other way to solve this problem?

mau5 · Answer

You can see that  15+4\sqrt{14} = 15+2*2*\sqrt{2}\sqrt{7}

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

=  (2\sqrt{2}+\sqrt{7})^2

Similarly. the same for the other part and after the square root of the terms, you get  (2\sqrt{2}+\sqrt{7})  +  (2\sqrt{2}-\sqrt{7})  = 2*2\sqrt{2}^2  = 32

E.

Sash143 · Answer

put x= 15+4\sqrt{14} and y= 15-4\sqrt{14} ==> (\sqrt{x}+\sqrt{y})^2 ==> x+2\sqrt{xy}+y Substitute for x and y ===> 30+2\sqrt{225-224} ===> 32 option D

law258 · Answer

(a+b)^2 = a^2 + 2ab+ b^2

15+2+15 =  32

JeffTargetTestPrep · Answer

We can let

x = √(15 + 4√14) and

y = √(15 - 4√14).

Since (x + y)^2 = x^2 + y^2 + 2xy, we have:

x^2 = (15 + 4√14)

y^2 = (15 - 4√14)

and

2xy is

2√(15 + 4√14) x √(15 - 4√14)

2√

Since (15 + 4√14)(15 - 4√14) can be written as a difference of squares, we have:

2√

2√  = 2

Thus, the total value is:

(15 + 4√14) + (15 - 4√14) + 2 = 32

Answer: E

sasyaharry · Answer

Once you reach this step:

2\sqrt{15^2-16*14} .

You can further simplify  16 * 14 = (15+1) (15-1) = 15^2 - 1^2

2\sqrt{15^2-(15^2 - 1^2)} = 2

BrushMyQuant · Answer

↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧

https://www.youtube.com/watch?v=6tbY4vtheWM

We need to find the value of  (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2

This is of the form  (a + b)^2  and will be equal to  a^2 + 2ab + b^2

=>  (\sqrt{15+4\sqrt{14}}+\sqrt{15-4\sqrt{14}})^2  =  (\sqrt{15+4\sqrt{14}})^2 +2* \sqrt{15+4\sqrt{14}} * \sqrt{15-4\sqrt{14}} + (\sqrt{15-4\sqrt{14}})^2 
=  15+4\sqrt{14} + 2 * \sqrt{(15+4\sqrt{14}) * (15-4\sqrt{14})}+ 15-4\sqrt{14}

Now,  (15+4\sqrt{14}) * (15-4\sqrt{14})  is of the form (a+b) * (a-b) and will be equal to  a^2 - b^2

=> 30 + 2 *  \sqrt{ 15^2 - (4\sqrt{14})^2}  = 30 + 2*  \sqrt{225 - 16*14}  = 30 + 2 *  \sqrt{1}  = 30 + 2 = 32

So,  Answer will be E 
Hope it helps!

Watch the following video to learn the Basics of Roots

https://www.youtube.com/watch?v=5oAec-Y6kOQ

SirSanguine · Answer

Here is an approach which will help you to answer this in 30 seconds.
we have to use (a+b)^2 realise that a^2 + b^2 will give you 30(15+15). Now, we have +2ab ie an even number is to be added only one option is there of the type 30 + even number.
cheers!

AnkurGMAT20 · Answer

Bunuel , this one came in FE Mock 5. Please tag it.

Thank you

Bunuel · Answer

______________________
Added the tag. Thank you!

JoeAa · Answer

can you explain how  2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}=2\sqrt{15^2-16*14}  ?

Bunuel · Answer

2\sqrt{15+4\sqrt{14}}*\sqrt{15-4\sqrt{14}}=

=2\sqrt{(15+4\sqrt{14})(15-4\sqrt{14})}

Apply  (a+b)(a-b)=a^2-b^2 :

=2\sqrt{15^2 -(4\sqrt{14})^2} =
 =2\sqrt{15^2-16*14}

napolean92728 · Answer

Consider,  15  as  a  and  4\sqrt{14}  as  b  , upon simplifying you get,  2a + 2(a^2-b^2) = 30 + 2(225-224) = 32

bumpbot · Answer

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((15 + 4(14)^(1/2))^(1/2) + (15 - 4(14)^(1/2))^(1/2))^2

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