Radical 7 and 29

What is the value of [√(7+√29) - √(7-√29)]²?

-26

2√29

14-4√5

14

14+4√5

The square of a value must be nonnegative.
Eliminate A.

Rather than calculate, BALLPARK:
[√(7+√29) - √(7-√29)]²

≈ [√(7 + 5.5) - √(7 - 5.5)]²

= [√(12.5) - √(1.5)]²

≈ [3.5 - 1]²

≈ (5/2)²

= 25/4

≈ 6.

The correct answer choice must be close to 6.
B, D and E are all too big.

The correct answer is C.

Algebraic approach:
Let x=7 and y=√29.

Substituting x=7 and y=√29 into [√(7+√29) - √(7-√29)]², we get:
( √(x+y) - √(x-y) )²

= (x+y) + (x-y) - 2√((x+y)(x-y))

= 2x - 2√(x²-y²).

Substituting x=7 and y=√29 back into the resulting expression, we get:
= 2*7 - 2√(7² - 29)

= 14 - 2√20

= 14 - 4√5.

The correct answer is C.

Thank you very much GMATGuruNY!

Hi

Just start with the basic approach to attack this problem


(a-b)^2 = a^2 +b^2 -2ab - (i)
a= squareroot(7 + squareroot 29)
b= squareroot(7 - squareroot 29)

after applying the formula in equation (i)

= (7 + root29) + (7-root29) - 2squareroot((7+squareroot29)(7-squareroot29))
= 14 - 2squareroot(49-29)
= 14 - 2squareroot(20)
= 14 - 4squareroot(5)

Let me know if you still have doubts




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I'd do it this way.

Let a = √(7 + √29) and b = √(7 - √29).

We know that

(a - b)² =>

a² - 2ab + b²

Now let's plug our a and b into this equation:

(√(7 + √29))² - 2*(√(7 + √29))*(√(7 - √29)) + (√(7 - √29))²

Notice that the end terms add up nicely:

(√(7 + √29))² + (√(7 - √29))² = 7 + √29 + 7 - √29 => 14

So we've got at least 14. Now let's deal with the middle term. We have

- 2*(√(7 + √29))*(√(7 - √29))

Remember that

√(c + d) * √(c - d) =>

√((c+d)*(c-d)) =>

√(c² - d²)

Since we have c = 7 and d = √29, this gives us √(7² - √29²), or √(49 - 29), or √20, or 2√5. Since we're doing -2 * this, we have -2 * 2√5, or -4√5.

Adding the two pieces together gives us 14 - 4√5, and we're done!