Main post link -> http://pibyzero.blogspot.in/2013/05/q.html
Q. A natural number is greater than three times its square root by 4. Find the number.
let, the number be x.
hence according to the condition- x = 3√x + 4
x-4 = 3√x
x² - 8x + 16 = 9x
x² - 17x + 16 = 0
(x-16)(x-1) = 0
x = 16 or x = 1
But when the negative value of square root of 16 and that positive of 1 is taken the equation is not satisfied. WHY ?????????????? Where is the mistake ?
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Comments
Squaring introduces an extraneous root. For example, if one plugs 1 into x-4=3sqrt(x), one gets -3=3, which is clearly not true, but the statement resulting from squaring both sides would be true.
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But if the same question is asked in examination what should be the answer ?
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I'm not going to help you cheat. You can figure that one out for yourself.
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if we take negative value,we get complex roots to the equation x^2-5x+16=0 which contradicts to question,since it is a natural number.
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No I'm not talking about the negative value of the root but the negative value of square root since every positive number has two opposite roots [ ( \sqrt{1} ) = ± 1 and ( \sqrt{16} ) = ± 4 ]
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-4 does not satisfy the initial conditions of the question. The question clearly states we need only the square of a natural number. -4 is not a natural number. Hence, it is illogical, subject to the initial conditions of the question, to write the square root of 16 as -4.
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Take the solution 16[it is a natural OK]
according to condition-
x = 3√x + 4
LHS = 16
RHS = 3√16 + 4
but there are two values of √16
what to do about them????????
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when we square any equation with a variable....the no. of roots of...increases automatically.....in those some roots does not satisfies the equation.....
also if u check out....√(-4) = 4i
which is a complex no. and solutions can only be natural no....hence -4 doesn't satisfies
Hey You all are in misconception that it is a problem in any exam or homework. But it is not so . In fact it is a problem in the Question bank (whose problems are not forced to solve) in my algebra book[10 th std. algebra book of Maharashtra State Board of Secondary and Higher Education]. These problems are only for practice. But I had asked a suggestion from you on it because I was confused how to write the answer of same kind of confusing question if asked in real Exams. I hope you will concentrate about the question and not the asker's purpose.
Since it is not stated that the square root of the natural number should be positive or negative. I think 1 will also satisfy the equation. 1=31+4
⇒1=−3+4
1=1
which clearly satisfies the above conditions. So both 1 and 16 qualify as answers.
16 = +4 and not equal to -4.
consider y2 = 16
y = +16 or y = -16
So,
16 = 4 and - 16 = -4
Furthermore, for all real x, x is always positive. It is not a function that takes two values.
So + 16 can be only one number and a positive number.
This is also the reason why * x2 = | x| *