Subtle Subtraction
x 2 − 5 a x + 1 0 0 = 0 x 2 − 1 0 0 x + 5 a = 0
Find the sum of all possible values of x that satisfy the equations above. Note that a is an arbitrary constant.
The answer is 99.
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8 solutions
There is some ambiguity in this problem in that a is not constant. Two separate systems with different solutions. One system has a=20 and the other a=-101/5.
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I agree. The problem poster should include in the problem that a is an arbitrary constant.
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Ok ... I will add that! Thanks!
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@Nihar Mahajan – Hey!I don't understand the last step of your solution.Why have you subtracted 100 by 1?
Sorry , But I did not get your point. Can you please elaborate more?Thanks!
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He means that you haven't specified the fact that a is an arbitrary constant.
Really, it's a system of two equations in two unknowns, a and x . In this case, there turn out to be three solutions: a = − 1 0 1 / 5 , x = − 1 a = 2 0 , x = 5 0 − 2 0 6 a = 2 0 , x = 5 0 + 2 0 6 What the problem is asking for is the sum of the distinct values of x in the solutions.
The question does mention that a is an arbitrary constant when I answer. Arbitrary constant: a quantity of function that is introduced into the solution of a problem, and to which any value or form may at will be given, so that the solution may be made to meet special requirements.
Very nice problem :D
Can you explain to me what Vieta's Formula is ??
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Oh friend , Vieta's formula gives the relation between the roots and coefficients of a polynomial. To get information Please click here
Did you remember to verify that the three values of x so derived do indeed satisfy the equation? Sometimes in an equation you can deduce logically that x has one of a number of values, but this doesn't mean the converse (if x is one of these values, the original equation is satisfied) is necessarily true. For instance, this is a logically valid argument x = 2 0 ⇒ x 2 = 4 0 0 ⇒ x ∈ { − 2 0 , 2 0 } but nonetheless, x = − 2 0 doesn't satisfy the original equation. While it's hard to make the mistake with this particular equation, I have made similar mistakes with more complicated equations on Brilliant.
@Nihar Mahajan , could you tell me why did you subtract 1 from 1 0 0 in the end? If the sum of roots is 1 0 0 , then why 1 0 0 is not the answer. Please help me to remove my confusion.
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Because it does not include − 1 in the sum , which we have calculated separately.
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How is it not included?
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@Seeku Hhh – We have two cases : x = − 1 and 5 a = 1 0 0 . From the 1st case ( − 1 ) is a root. From the second case , sum of roots = 1 0 0 . So considering both cases , the sum of all the roots = 1 0 0 − 1 = 9 9 .
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@Nihar Mahajan – You mean that − 5 a x − 5 a + 1 0 0 + 1 0 0 x = 0 is a separate equation and x 2 − 1 0 0 x + 1 0 0 is separate?
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@Seeku Hhh – I don’t get how it’s separate . I’m guessing when you substitute random equation, you somehow get different roots
Is value of "a" equal only 20?
Substitute value of a from one equation into another to get
x 3 − 9 9 x 2 + 1 0 0 = 0
The required sum = -(-99)/1 = 9 9 .
x^2 - 5 a x + 100 = 0 ..................... (1)
x^2 - 100 x + 5 a = 0 ......................(2)
From Eq (2) we get
5 a = 100 x - x^2
Substituting in Eq (1) we get
x^3 - 99 x^2 + 100 = 0
So
Sum of roots = -(-99)/1 = 99
What an alternative!
very simple one
great thinking
I'll add a subtle point too!
Let, x = − 1 Put this value in one of the equations. You'll get 1 + 5 a + 1 0 0 = 0 or, 5 a = − 1 0 1 or, a = − 5 1 0 1 Turns out the value of a is not that arbitrary. It just have two values. a = 2 0 , − 5 1 0 1
Am I right?
By the second equation
5a=100x - x^2
Plugging this value for 5a into the first equation we get
x^3 - 99x^2 + 100 = 0
By Vieta, the sum of the roots is 99.
Well, simply put the value of 5a from the second equation into the first one........you will get a cubic equation in x.......Then, by Vieta's the sum of all possible values of x is 99.....!!
Nice solution. But I did it the other way. This was an easy one.
get 5a=x+100/x from first and substitute in second.A cubic is obtained where applying Vieta's formula sum of all values of =99
Subtract the second equation from the first to get :
− 5 a x − 5 a + 1 0 0 + 1 0 0 x = 0
⇒ − 5 a ( x + 1 ) + 1 0 0 ( x + 1 ) = 0
⇒ ( x + 1 ) ( 1 0 0 − 5 a ) = 0
This gives us a solution x = − 1
But , If 5 a = 1 0 0 , we see that the both equations become same!
So , when a = 2 0 , Substitute it in equation 1 to give us:
x 2 − 1 0 0 x + 1 0 0
By Vieta's formula , Sum of roots = 1 0 0
Hence , sum of all values of x = 1 0 0 − 1 = 9 9