Radical System
a + b = 7 b + a = 1 1
If a and b are real numbers that satisfy the equation above, what is the value of a and b respectively?
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12 solutions
My approach is slightly simpler, without having to notice the "nice" factorization that is implied by the answer.
I substituted the second equation into the first, to get x + ( 1 1 − x 2 ) 2 = 7 , which gives us ( x − 3 ) ( x 3 + 3 x 2 − 1 3 x − 3 8 ) = 0 . Hence x = 3 is a solution.
For the cubic equation, there are 2 negative roots (which we can ignore), and the third root is about 3 . 5 8 4 4 . This gives us a ≈ 1 2 . 8 4 7 9 and b ≈ 7 . 4 1 5 6 . However, this doesn't satisfy the second equation because we actually have − 7 . 4 1 5 6 + 1 2 . 8 4 7 9 = 1 1 .
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Calvin, I propose a new name for one approach to solving problems. We already have "bashing" which denotes solving problems using brute force. How about "feigned prescience" to denote solutions that clearly depend on already knowing what the answer is? I see a lot of solutions of that nature, and not just here in Brilliant (some of which I'm guilty of!).
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True. Sometimes I start with "This solution is based on already knowing / guessing what the answer is ...". This happens when the solution is "by magic" and hard to find the motivation.
Often, I try and drill into the solution further, and highlight certain aspects which could explain a more intuitive approach. This isn't always possible, in part because Problem Solving often has a "Ah-ha" moment that is hard to quantify down.
no thats wrong..u have substituted the value of (3-x) with (x-3)..u cant do that If (x-3) = (2-y)(2+y), then (3-x) = - (2-y)(2+y)
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Do you know who you're talking to?
Interesting approach. There could be other solutions from the second factor right? So you have to explain why it is non-zero.
Or you could just check all 4 possible solutions..
the second to last equation needs a negative sign on the right side.
Brilliant! After the x^2 =a and y^2=b substitutions, simply plot a few points for the y functions y1 = +/- (7 - x)^.5 and y2 = (11- x^2). I.e., let x =0,1,2,3... At x=3, y1=y2=2. Hence, x^2=9=a, and y^2=4=b. That's a lot easier than trying to solve, a^4 - 44 a^3 +712 a^2 - 5017*a +12996 I'm ashamed to say.
How are we able to make those substitutions? (still learning)
For that rearrengement, you need to know the answer. Other type arrengements wont give any simplementation.
how is it that you didnt change the sign (y−2)=(3−x)(3+x)
Which leads to
(y-2)=(2-y)(2+y)(3+x)(y−2)=(2−y)(2+y)(3+x)
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what if we took x=-a or y=-b
Respected Sir, there is a little mistake .. there sould be (y-2) instead of (2-y), after Which leads to..
I am most unsatisfied. To me, trial and error is picking possible solutions, then seeing which one is correct. Even the algebraic solutions calculate 4 possible answers, then pick (through trial and error) the correct one. It appears, given the stipulation that the problem disallowed trial and error, that the correct answer to the problem is "not able to be solved."
Cool mahnn.. You nailed it.. :)
kindly elaborate more>>>>>???
Only option no. 3 has two numbers which are squares. Only perfect squares can give a whole number. Its that easy.
What if they didn't have the choices?
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Then you need to do the workings. You need to save time in exams, so look for the reasonable and LOGICAL way, if you want to find an answer quickly!
PS : It's always the survival of the fittest in the world. It's not wrong if you have a trick up your sleeve to protect you!(I hope you understood what I am trying to say!)
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No need of much calculations just cheak with options. Only option 3 is matching
What if they didn't have the choices?
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Actually, even without choices, guessing and enumeration still is the fastest way to solve this. We establish that a and b should be perfect squares since both sums are whole numbers. Looking at the first equation, we can immediately see, by listing all possible addends of 7 (7 + 0, 6 + 1, 5+ 2,...), that only 1 and 4 are perfect squares; hence unknown a must be either of those two. Suppose a is 1, we have to add 6 to it to arrive at 7, which contradicts our non-negotiable observation that both a and b must be perfect squares. Therefore a must be 4. 4 + 3 = 7 and hence b = 9.
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a is 9 and b is 4 not a is 4 if a is 4 b will be five bec 2+5 =10
Spelling of check is incorrect
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Firstly, isolate b in a + b b = 7 − a Then subsitute it into the equation b + a = 1 1 7 − a + a = 1 1 Its a square root nested in a square root! First, subtract a 7 − a = 1 1 − a Then square both sides. 7 − a = a 2 − 2 2 a + 1 2 1 One square root eliminated. Subtract 7 from both sides, − a = a 2 − 2 2 a + 1 1 4 Square both sides again! a = a 4 − 4 4 a 3 + 7 1 2 a 2 − 5 0 1 6 a + 1 2 9 9 6 Now, SOLVE THE EQUATION a = a 4 − 4 4 a 3 + 7 1 2 a 2 − 5 0 1 6 a + 1 2 9 9 6 Its tricky, and it has 4 solutions (by the a 4 as you can see), so I recommend using a graphing calculator and find the solutions. They are 9, 7.86869..., 12,84813..., and 14.28319... Verifying the solutions, only 9 is true. Thus, we got a = 9 Now, find b 9 + b = 7 b = 4
So, the solution set is { a = 9 , b = 4 }
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Backtracking from the given choice is the quickest way to find the solution, If problem did't have choice of solution we need to go with substitution method. Also there is one more hint in the problem 'a' and 'b' mostly be a perfect square.
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a and b should be integer after extraction of a root.
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I have added a better solution below please check it out
It's easy to confirm via substitution in the first equation !
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9 square is 3 and 4 square is 2 Then A value is 9 B value is 4
What if they didn't have the choices?
binomial theorm 71\8
Let α = a and β = b . We get α + β 2 = 7 and α 2 + β = 1 1 . It is clear that for real roots α ∈ [ 0 , 1 1 ] and β ∈ [ 0 , 7 ] .
By substituting α = 7 − β 2 into α 2 + β = 1 1 we get the polynomial β 4 − 1 4 β 2 + β + 3 8 = 0 Hoping for rational solutions, we use the rational root theorem . This implies that any rational root β has to be an integer divisor of 38, of which only 1 and 2 are in the allowed range. Checking for β = 2 , we see it is a solution indeed, because 1 6 − 5 6 + 2 + 3 8 = 0 . Now b = 4 and a = 9 follow directly.
I square that a = 9. But. There is a way to make (x + 1). and if we add 4. we will have a continuing process.
hello, we can solve this equation just by assuming... sq. root of 'a'+'b'=7, 'a'+sq. root of 'b'=11 these equations apply that a and b both are squares of a particular no. so, lets assume a=9 and b=4(as both are squares of a particular no.) so lets see if they satisfy the equation or not.. sq. root of 9=3 and sq. root of 4=2...., 3+4=7 9+2=11 here both of the equations are been satisfied.. Therefore, a=9 and b=4
First make the substitutions x 2 = a and y 2 = b . Then we have
x + y 2 = 7
y + x 2 = 1 1
Re-arranging both, we have
( x − 3 ) = ( 2 − y ) ( 2 + y )
( y − 2 ) = ( 3 − x ) ( 3 + x )
Which leads to
( y − 2 ) = ( 2 − y ) ( 2 + y ) ( 3 + x )
That is
( y − 2 ) ( 1 − ( 2 + y ) ( 3 + x ) ) = 0
Being that y − 2 is one of the factors, b = 4 , and from that a = 9