Question

Find a negative real number such that the square of the sum of a number and 5 is equal to 48

Answer 1

Let x = negative real number  ⇒x<0

from the statement above, we can generate an equation: 
(x + 5)² = 48
$\sqrt{(x+5)^{2} }$= $\sqrt{48}$  
⇒ eliminate the square by getting the square root on both sides

$\sqrt{48} = \left \{ {{=4 \sqrt{3} } \atop {=-4 \sqrt{3} }} \right.$
⇒ the perfect square of a real number has one positive real number and a negative real number

transposing 5 to other side, we will arrive at two (2) values for x:

$x_{1}$ = -5 - 4√3    = -11.928
$x_{2}$ = -5 + 4√3   =  1.928

Since we are only looking at the negative real number, our answer will be -11.928, also equal to -5 - 4√3.