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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

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TopNewestSquaring 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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I was asking for the steps and reasons and not the solutionLog in to reply

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. \(1=3\sqrt{1}+4\)

\(\Rightarrow 1=-3+4\)

\(1=1\)

which clearly satisfies the above conditions. So both 1 and 16 qualify as answers.

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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.

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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

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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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which are natural numbers) see the two ways...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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\(\sqrt{16}\) = +4 and not equal to -4.

consider \(y^{2}\) = 16

y = +\(\sqrt{16}\) or y = -\(\sqrt{16}\)

So,

\(\sqrt{16}\) = 4 and - \(\sqrt{16}\) = -4

Furthermore, for all real

x, \(\sqrt{x}\) is always positive. It is not a function that takes two values.So + \(\sqrt{16}\) can be only one number and a

positivenumber.This is also the reason why

* \(\sqrt{x^2}\) = | x| *Log in to reply