# What's wrong in it ???????????????

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 ?

Note by Rushikesh Jogdand
5 years, 6 months ago

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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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

agreed.........:)

- 5 years, 6 months ago

It's called a hypothetical, Bob. No need for the attitude.

- 5 years, 6 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.

- 5 years, 6 months ago

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. I was asking for the steps and reasons and not the solution

- 5 years, 6 months ago

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.

- 5 years, 6 months ago

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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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.

- 5 years, 6 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 ]

- 5 years, 6 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.

- 5 years, 6 months ago

It doesn't says square root must be natural. It says the numbers should be natural...

- 5 years, 6 months ago

No. You are again misunderstanding my side. I am telling that there is confusion between square root os the solutions. The solutions are 16 and 1 (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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

$$\sqrt{16}$$ = +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 positive number.

This is also the reason why * $$\sqrt{x^2}$$ = | x| *

- 4 years, 11 months ago