Hello all,
I am having difficulty simplifying this question that I found within my first Manhattan Prep CAT:
Any help will be much appreciated.
Thank you,
mcdo0351
Radical 7 and 29
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The square of a value must be nonnegative.What is the value of [√(7+√29) - √(7-√29)]²?
-26
2√29
14-4√5
14
14+4√5
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.
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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
[/list]
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
[/list]
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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!
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!