Question

A natural number is greater than three times its square root by 4. Find the number.
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Answer 1

Answer:

The natural number is 2.

Answer 2

Let the number be 'x'

☆a/q to question:-

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

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

$squaring \: both \: sides$

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

we know that

(a-b)²=a²-2ab+b²

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

$x {}^{2} - 8x - 9x + 16 = 0$

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

$(x - 16)(x - 1) = 0$

$x = 16 \: or \: x = 1$

But x =1 isn't possible

so,

$\mathbf \red{x = 16}$