Question

a natural number is graeter than three time its square root by 4 find the number
​

Answer 1

$\bold{\large{\underline{\tt{Answer}}}}$

$\bold{\large{\underline{\tt{\blue{The\:Natural\:Number\:is\:16}}}}}$

Given :

• A natural number is graeter than three time its square root by 4.

To Find :

• The Number

Solution :

Let the natural number be x.

$\bold{\underline{\underline{Acc.to\:the\:question:}}}$

➟ $\mathtt{x=\:3\:{\sqrt{x}\:\:+\:4}}$

➟ $\mathtt{x-4\:=\:3\:{\sqrt{x}}}$

Squaring both sides,

➟ $\mathtt{(x-4)^2\:=\:3^2\:({\sqrt{x})^2}}$

➟ $\mathtt{x^2\:-\:2\:\times\:x\:\times\:4\:+\:16\:=\:9x}$$\sf{\underbrace{Using\:(a-b)^2}}$

➟ $\mathtt{x^2\:-\:8x\:+16\:=\:9x}$

➟ $\mathtt{x^2\:-8x\:-9x\:=\:-16}$

➟ $\mathtt{x^2\:-17x\:+16\:=\:0}$

Now, the equation formed is a quadratic equation which we can easily solve and find the value of x.

➟ $\mathtt{x^2\:-16x-x\:+\:16\:=\:0}$

➟ $\mathtt{x(x-16)\:-1(x-16)\:=\:0}$

➟ $\mathtt{(x-16)\:\:\:(x-1)\:\:=\:0}$

➟ $\mathtt{x-16\:=\:0\:\:OR\:\:x-1=0}$

➟ $\mathtt{x=16\:\:OR\:\:x=1}$

We have two positive values of x. Though from the given statement we can infer that x = 1 won't satisfy the required condition.

But still let's verify.

When, x = 1 :

➟ $\mathtt{x=\:3\:\times\:{\sqrt{x}\:+4}}$

➟ $\mathtt{1\:=\:3\:\times\:{\sqrt{1}\:+4}}$

➟ $\mathtt{1=3\:\times\:1\:+\:4}$

➟ $\mathtt{1\:=3\:+\:4}$

➟ $\mathtt{1=7}$

LHS ≠ RHS.

Therefore, x = 1 is neglected.

When, x = 16 :

➟ $\mathtt{x=\:3\:\times\:{\sqrt{x}\:+4}}$

➟ $\mathtt{16\:=\:3\:\times\:{\sqrt{16}\:+4}}$

➟ $\mathtt{16=3\:\times\:4\:+\:12}$

➟ $\mathtt{16\:=\:12+4}$

➟ $\mathtt{16=16}$

LHS = RHS.

Hence, x = 16 is the required natural number.

Answer 2

Hey Pretty Stranger!

$\textbf{\underline{\underline{Given:}}}$

• A natural number is greater than 3 times its square root by 4

$\textbf{\underline{\underline{To\:Find:}}}$

• The natural number

$\textbf{\underline{\underline{Solution:}}}$

Let the natural number be x

According to the Question :

$\sf { x =3 \sqrt{x} + 4 }$

$\sf \:x - 4 = 3 \sqrt{x}$

Ek se Bhale Do, Let's square both the sides. :D

$\sf \: ( {x - 4})^{2} = ({3 \sqrt{x} )}^{2}$

$\sf \: {x}^{2} - 8x + 16 = 9x$

$\sf \: {x}^{2} - 8x - 9x = - 16$

$\sf \: {x}^{2} - 17x = - 16$

$\sf \: {x}^{2} - 17x + 16 = 0$

$\sf \: {x}^{2} - 16x - x + 16 \: = 0$

$\sf \: x(x - 16) - 1(x - 16) = 0$

$\sf \: (x - 1) \: (x - 16)$

The two required values we have got is 16 and 1 . It's clear that 1 can't be that natural number.

HENCE,

$\large { \purple {\boxed {\boxed{ \sf \:16 }}}} \: - \sf Required \: Natural \: No.$