If (8 + 3*21^(1/2))^(1/3) + (8 - 3*21^(1/2))^(1/3) = x, what is the va

Question

If  \sqrt {8 + 3\sqrt{21}} + \sqrt {8 - 3\sqrt{21}} = x , what is the value of x?

A. 1/2
B. 1
C. 3/2
D. 2
E. 3

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nick1816 · Accepted Answer

\sqrt {8 + 3\sqrt{21}} + \sqrt {8 - 3\sqrt{21}} = x

{8 + 3\sqrt{21}} = (a+b)

{8 - 3\sqrt{21}} = a-b

x = \sqrt {a+b} + \sqrt {a-b}

We know that

(c+d)^3 = c^3+d^3 +3cd(c+d)

Hence,  x^3 = (a+b)+(a-b) +3 \sqrt {a^2-b^2} *  {a+b} + \sqrt {a-b}]

a^2-b^2 = 8^2 -  ^2 = 64-189 = (-125)= (-5)^3

x^3 = 2a +3 *-5* x

x^3 = 16-15x

Use options

x=1

GMATinsight · Answer

I am just approximating because I love it.

\sqrt {8 + 3\sqrt{21}} + \sqrt {8 - 3\sqrt{21}} = x 4.5^2 = 20.25 i.e. √21≈ 4.6 i.e. \sqrt {8 + 3*4.6} + \sqrt {8 - 3*4.6} = x i.e. \sqrt {22} + \sqrt {-5.8} = x because 2^3 = 8 and 3^3 = 27 therefore 2.8^3≈22 i.e. 2.8 + -1.8 ≈ 1 Answer: Option B

yashikaaggarwal · Answer

GMATinsight  you might have to edit {-4} above, (8-3*4.6) => -5.8 or -6

Posted from my mobile device

GMATinsight · Answer

yashikaaggarwal Thank you. Edited.

ScottTargetTestPrep · Answer

Solution:

We can let a = 3^√(8 + 3√21) and b = 3^√(8 - 3√21). Notice that a^3 = 8 + 3√21, b^3 = 8 - 3√21 and a*b = 3^√(64 - 9(21)) = 3^√(-125) = -5. Now let’s cube both sides of the equation:

(a + b)^3 = x^3

a^3 + 3a^2*b + 3a*b^2 + b^3 = x^3

a^3 + 3ab*a + 3ab*b + b^3 = x^3

8 + 3√21 + 3(-5)a + 3(-5)b + 8 - 3√21 = x^3

16 -15(a + b) = x^3

But a + b = x, so we have:

16 - 15x = x^3

From the answer choices, we see that x must be 1 in order to satisfy the equation.

Answer: B

chetan2u · Answer

Are we supposed to know algebraic formulas used here - may be no, or are we supposed to know  \sqrt{21} , may be no.
So if we ever get such a question, we would be required to approximate.

Now, as far as the choices are concerned,  we can eliminate the fractions  because the question does not ask for approximate value and we can not get a terminating decimal here, So surely somewhere square root must be getting cancelled. So A and C can be eliminated.
 E can be eliminated  as we can see one of the term will be negative and other negative, and the larger term itself will be smaller than 3 or  \sqrt {27} .

We can also eliminate D , that is x=2 as by quick approximation, we can see the answer is closer to 1 than to 2.
Now  \sqrt{16}<\sqrt{21}<\sqrt{25} , so  x<\sqrt {8 + 3\sqrt{25}} + \sqrt {8 - 3\sqrt{25}} =\sqrt {8 + 3*5} + \sqrt {8 - 3*5} =\sqrt {23} + \sqrt {-7}  ~3-2=1....so x cannot be 2

But if we want to get a bit more closer in our approximation 
 \sqrt{21}=\sqrt{3*7}=\sqrt{3}*\sqrt{ 7}=1.7*2.6=4.4

\sqrt {8 + 3\sqrt{21}} + \sqrt {8 - 3\sqrt{21}} =\sqrt {8 + 3*4.4} + \sqrt {8 - 3*4.4}  
 \sqrt {21} + \sqrt {-5}  Surely less than 3-1 or <2.

Regor60 · Answer

Set W= 8+3(21^1/2) and Y= 8-3(21^1/2)

So W^1/3 + Y^1/3 = X

Cube both sides😬 and collect terms

W+Y + 3(w^1/3)(y^1/3)(w^1/3 + y^1/3) = x^3

W+y= 16 and (w^1/3)(y^1/3) = (-125^1/3) = -5 and substitute back in X

16-15X = X^3

X(X^2+15) = 16

1(1+15) = 16, so X=1

Posted from my mobile device

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If (8 + 321^(1/2))^(1/3) + (8 - 321^(1/2))^(1/3) = x, what is the va

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