Greatest Confusion!

Algebra Level 3

A = 8 + 22 B = 1 + 29 C = 12 + 18 D = 10 + 20 \begin{aligned} A &=& \sqrt 8 + \sqrt{22} \\ B &=& \sqrt 1 + \sqrt{29} \\ C &=& \sqrt{12} + \sqrt{18} \\ D &=& \sqrt{10} + \sqrt{20} \end{aligned}

Using the above information, which of these numbers is the largest in value?

B B A A C C D D

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7 solutions

Pi Han Goh
Feb 25, 2015

Hint: consider the square of each of the expressions.

Moderator note:

Good. Although a further elaboration would be better.

Ya the same here . It will give only single root over and then it will be easy to find the value !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Abhisek Mohanty - 6 years, 2 months ago
Chew-Seong Cheong
Apr 22, 2015

For x + y \sqrt{x}+\sqrt{y} with x + y = c x+y = c , a constant, then smaller x y |x-y| , the larger the x + y \sqrt{x}+\sqrt{y} . Check out the following table.

x y x y x + y 12 18 6 7.70674230225704 10 20 10 7.63441361516796 8 22 14 7.51884288456962 1 29 28 6.3851648071345 \begin{array} {rrrl} x & y & |x-y| & \sqrt{x} + \sqrt{y} \\12 & 18 & 6 & 7.70674230225704 \\ 10 & 20 & 10 & 7.63441361516796 \\ 8 & 22 & 14 & 7.51884288456962 \\ 1 & 29 & 28 & 6.3851648071345 \end{array}

Maximum is when x = y = 15 x=y=15 .

Moderator note:

This is no different from a brute force method. Can you provide an alternative solution for this question (other than squaring the expression)? Hint: calculus.

Lol... Max can be infinite! Rest of your logic looks fine....

Jai Mahajan - 6 years, 1 month ago

Log in to reply

How can the max be infinite for the set where X + Y = 30?

Robert Montgomery - 6 years, 1 month ago

he assumed a function f(x,y) defined on [0:30] such that x+y = 30 for every x and y chosen and that's why maximum is happening at x = y =15 because you'd get 225 when squaring the radicals

Radinoiu Damian - 6 years, 1 month ago

I'm still not getting how you came to these figures. why would you be factoring in the x-y

michael bye - 6 years, 1 month ago

Thank you sir :-)

Jyo Moy - 6 years, 1 month ago
Marilyn Bretscher
Apr 24, 2015

f ( x ) = 15 x + 15 + x = 30 + 2 225 x 2 f(x)=\sqrt{15-x}+\sqrt{15+x}=\sqrt{30+2\sqrt{225-x^2}} ( for 0 x 15 0\leq{x}\leq15 )

is decreasing. Thus we choose the smallest available x x , namely, x = 3 x=3 in C \boxed{C} .

Moderator note:

You need to prove that it's a decreasing function in the said interval.

Can you explain why the function on the left is equal to the function on the right? Thank you!

Minh Tran - 6 years, 1 month ago
Md Faisal Akbar
Apr 24, 2015

Square all the term then make this form (a+b)^2= a^2+b^2+2ab then compare

This is the answer that I look for. Thank you!

Minh Tran - 6 years, 1 month ago

yeah i totally get it. i was doing this at 2am and didnt realize that the calculator i was using wa only giving me the root of the 2nd number and not totaling the entire problem lol

michael bye - 6 years, 1 month ago
Manish Sharma
Apr 24, 2015

12 *18=216 largest

10*20=200

8*22=176

1*29=29 smallest

And so the larger it is, the larger the sum of the square root?

Minh Tran - 6 years, 1 month ago
Arian Tashakkor
Apr 27, 2015

A 2 = 30 + 176 A^2=30+\sqrt{176}

B 2 = 30 + 29 B^2=30+\sqrt{29}

C 2 = 30 + 216 C^2=30+\sqrt{216}

D 2 = 30 + 200 D^2=30+\sqrt{200}

now comparing A 2 , B 2 , C 2 , D 2 A^2 , B^2 , C^2 , D^2 we can get that C C has the highest value.

Jason Myttas
Apr 24, 2015

What I thought is that as the number insider a root is larger, the difference of the value of the roots is smaller. For example, the difference between \sqrt{2} and \sqrt{3} is much bigger than \sqrt{100} and \sqrt{101} . Therefore we look at the first set of numbers that have the smaller values inside the square root, 1, 8, 10 and 12 and pick the biggest one, as the difference among the square roots of the bigger set of numbers, 18, 20, 22 and 29, will be smaller and will not affect the outcome as much.

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