Let x = negative real number ⇒x<0
from the statement above, we can generate an equation:
(x + 5)² = 48
[tex] \sqrt{(x+5)^{2} }[/tex]= [tex] \sqrt{48} [/tex]
⇒ eliminate the square by getting the square root on both sides
[tex] \sqrt{48} = \left \{ {{=4 \sqrt{3} } \atop {=-4 \sqrt{3} }} \right. [/tex]
⇒ 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:
[tex] x_{1} [/tex] = -5 - 4√3 = -11.928
[tex] x_{2} [/tex] = -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.
