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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 ?

## 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. – Bob Krueger · 4 years, 3 months ago

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– Rushikesh Jogdand · 4 years, 3 months ago

But if the same question is asked in examination what should be the answer ?Log in to reply

– Bob Krueger · 4 years, 3 months ago

I'm not going to help you cheat. You can figure that one out for yourself.Log in to reply

– Aditya Parson · 4 years, 3 months ago

How do you infer that he is trying to cheat? Please understand the argument he has proposed more clearly.Log in to reply

– Riya Gupta · 4 years, 3 months ago

agreed.........:)Log in to reply

– Tim Ye · 4 years, 3 months ago

It's called a hypothetical, Bob. No need for the attitude.Log in to reply

– Bob Krueger · 4 years, 3 months ago

You never know any more. I've seen too many of these discussions say "URGENT" or other key words like that which give me the impression they need it for an exam or school.Log in to reply

I was asking for the steps and reasons and not the solution– Rushikesh Jogdand · 4 years, 3 months agoLog 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. – Aditya Parson · 4 years, 3 months ago

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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. – Rushikesh Jogdand · 4 years, 3 months ago

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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 – Riya Gupta · 4 years, 3 months ago

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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. – Sri priya Prerna · 4 years, 3 months ago

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– Rushikesh Jogdand · 4 years, 3 months ago

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 ]Log in to reply

– Aditya Parson · 4 years, 3 months ago

-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.Log in to reply

– Rushikesh Jogdand · 4 years, 3 months ago

It doesn't says square root must be natural. It says the numbers should be natural...Log in to reply

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???????? – Rushikesh Jogdand · 4 years, 3 months ago

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– Aditya Parson · 4 years, 3 months ago

You take the values and separately solve them. One will satisfy the condition the other wont, simple.Log in to reply

\(\sqrt{16}\) = +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| *– Samarth M.O. · 3 years, 8 months agoLog in to reply