How careful are you with equations?
( x − 1 0 ) 2 + ( x − 1 5 ) 2 + ( x − 2 5 ) 2 = 0
How many distinct values of x satisfy the above equation ?
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5 solutions
Moderator note:
It would be better to clarify why it doesn't have real roots.
I feel very disappointed: The problem is incomplete as long as you don't state that x doesn't necessarily has to be a real number. You fool your audience by using a symbol, x, that is used 90% or more of times for real numbers, as the preferred variable to show that you admit complex solutions is to use z. So, IMHO, you are a cheater. Ok, you have reason, 2 solutions, you got me. But you're a cheater. Do you feel more intelligent now? cheers! Enjoy your conquest
:-(
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I disagree - the whole essence of the puzzle is that you have to realise for yourself that there may be complex roots.
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But there are always 2 roots to a quadratic. Just occasionally they are not distinct. i agree it's a poor question.
I do not agreee at all. Now you're suppossing x is a complex number. What if x ∈ Z / 5 Z , like other guy says? You will say: "wait, x usually means real number, not an element of these less common modular rings/fields/groups". That's my point: x is usually used for a real number. Using x instead of z is trying to trick your audience.
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@Jose Torres Zapata – Tricking your audience is sometimes the whole point of a puzzle. At the end of the day, x is just an arbitrary variable name. Usually it represents a real number, but there's no rule of maths or mathematical notation that says it absolutely must. Indeed, I think you'll find it's used fairly frequently in abstract algebra contexts.
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@Stewart Gordon – It is not about getting tricked, it is just about guessing and there is no point in doing maths if you want your audience to have guesses. Establish a proof that there should be two solutions to your equation no more no less, the way its formulated, and then we ll see. Good luck for that!
@Stewart Gordon – So, you want freedom for variables and notation. So be it. See, you've taken for granted that - means substraction in the usual sense!!. Let A be the set formed by real numbers 0, 10, 15, 25. Let us define operation - by a-b=0 for all a,b in A. Let us define taking square as the usual real number definition, and notation, multiply two times the base. Then the posed equation has 4 solutions. So the answer is 4. Usually, x means real, and - means substraction of real numbers, but hey! if you are tricking your audience, there is no place you should stop. Perphaps 80%+ of people got fooled by x not being real. Write the same problem with a z instead. Would it be the same number of wrong answers? I'll bet it wouldn't. With usual meaning of - and product, square, and = sign, if you take x ∈ Z / 5 Z there is only 1 solution, x=0. So, the answer would be 1.
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@Jose Torres Zapata – So, the "problem" is that the problem is no properly stated.
100% agree... because instead of "x" he could have used "i"
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I think the most common variables to show "I admit complex solutions" is w, or z. At least, that's what I've seen 90% I looked into any math book
Stupid problem. There is no point in trying to make people guess what is the question, just state PROPERLY or make real problems... Or stop doing mathematics maybe
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How is the question not stated properly?
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Because it's not clear what type of number x is supposed to represent.
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@Ron Sperber – I think it's implicit in the absence of any statement to the contrary that all kinds of numbers - positive and negative, integral and fractional, rational and irrational, real and complex, qualify as values satisfying the equation should they satisfy it.
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@Stewart Gordon – So then can x be a hyperreal number? a surreal number? an octonian? What if x is an element in some ring?
To fully state a problem asking for solutions, you must always state the space in which the solution should be found. There is no reasonable definition of 'all kinds of numbers.'
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@Kevin Driscoll – Yes, there is that problem. Unfortunately, I can't see a way to allow for this without destroying the essence of the problem, namely requiring the solver to think outside the box. :(
@Stewart Gordon – And in Z / 5 Z ? And what about the quaternions ?? "All kinds of number" has no meaning... In maths at least. Even if all is in italics !
In Z / 5 Z for example, 0 is solution, so your answer is wrong as well, there are (way) more than two "numbers" satisfying this equation.
couldn't agree more.
Among a^{2}, (a+5)^{2} and (a+15)^{2}, in worst case only one can be equal to zero and rest are strictly greater than zero. Hence, no real roots.
The given equation can be written as ( x − 1 0 ) 2 + ( x − 1 5 ) 2 + ( 2 5 − x ) 2 = 0 which is of the form a 2 + b 2 + c 2 ∴ x 2 = 2 [ ( x − 1 0 ) ( x − 1 5 ) + ( x − 1 5 ) ( 2 5 − x ) + ( x − 1 0 ) ( 2 5 − x ) ] = 2 [ 5 0 x − 4 7 5 − x 2 ] 3 x 2 − 1 0 0 x + 9 5 0 = 0
The above is a quadratic equation with D = 0 . Hence no. of distinct values of x= 2
Moderator note:
Nicely done. Although a suitable substitution would simplify the calculation by a little bit.
Sir your solution may be correct but i do not understand how come sum of three positive be zero. and if it is zero, each positive number has to be individually zero. so in that way X will have three values i.e. 10,15,25.
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The equation has imaginary roots and hence it is possible for the the equation to be true. No real values fulfill the given condition. I hope you get the point.
D= (-100)^2 - (4 3 9.5 100)=100^2 - 114 100 <0 =>roots of the eqn are complex. Thus the equation have two distincts imaginary solutions.
Nicely done! I made a silly error ⌢ ¨
Didn't read the question carefully LOL.
It's just a quadratic equation. By the so-called fundamental theorem of algebra, there are two roots. This may be a repeated root, or it may be two distinct roots. So the question is one of whether the roots are distinct.
Moreover, given a polynomial equation with only real coefficients, any non-real roots are necessarily in complex conjugate pairs. Therefore any repeated root is necessarily a real number. However, if x is real, then each of the terms (as the equation is written) is necessarily non-negative, and no more than one can be zero, therefore this would force the LHS to be positive. But since the RHS is 0, x cannot be real, therefore the roots are a complex conjugate pair.
Very nicely reasoned!
What on earth makes this a level 4 question!! .. I spent a fair amount of time questioning my own answer, questioning the meaning of distinction
Simplify [(-25 + x)^2 + (-15 + x)^2 + (-10 + x)^2]
950 - 100 x + 3 x^2
which indicates the highest power of x is 2 so the solution will be 2
A real quadratic equation without real solutions has 2 complex conjugate solutions.