Answer:
Option B is correct.
u= x+4
Step-by-step explanation:
Solve the equation:
[tex](x+4)^2-3(x+4)-3 = 0[/tex]
Using substitution:
Let u = x+4
then;
[tex]u^2-3u-3=0[/tex] ....[1]
For a quadratic equation [tex]ax^2+bx+c = 0[/tex] ....[2], then the solution is given by:
[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
On comparing equation [1] and [2] we have;
a = 1 , b = -3 and c = -3
then;
[tex]u= \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-3)}}{2(1)}[/tex]
⇒[tex]u= \frac{3 \pm \sqrt{9+12}}{2}[/tex]
⇒[tex]u= \frac{3 \pm \sqrt{21}}{2}[/tex]
Substitute u = x+4
then;
⇒[tex]x+4= \frac{3 \pm \sqrt{21}}{2}[/tex]
Subtract 4 from both sides we have;
[tex]x= \frac{3 \pm \sqrt{21}}{2} - 4= \frac{3 \pm \sqrt{21}-8}{2}[/tex]
⇒[tex]x= \frac{3 + \sqrt{21}-8}{2}= \frac{-5 + \sqrt{21}}{2}[/tex]
and
[tex]x= \frac{3 - \sqrt{21}-8}{2}[/tex]
⇒[tex]x= \frac{-5 - \sqrt{21}}{2}[/tex]
Therefore, the solution for the given equation are,
[tex]x= \frac{-5 - \sqrt{21}}{2}[/tex] , [tex]\frac{-5 + \sqrt{21}}{2}[/tex]
