Question

a natural number is greater than thrice it's square root by 4. Find the number

Answer 1

Let the natural no. be x

according to the question

$x - \sqrt{3x } = 4$
squaring both the sides

${x}^{2} - 3x = 4$
${x}^{2} - 3x - 4 = 0$
${x}^{2} - (4 - 1)x - 4 = 0$
${x}^{2} - 4x + x - 4 = 0$
$x(x + 1) - 4(x + 1) = 0$
$(x + 1)( x - 4) = 0$
so

$x = - 1$
OR

$x = 4$