MATH SOLVE

5 months ago

Q:
# Solve (x + 4)2 β 3(x + 4) β 3 = 0 using substitution. u = Select the solution(s) of the original equation.

Accepted Solution

A:

Answer:[tex]\dfrac{-5-\sqrt{21}}{2},\ \dfrac{-5+\sqrt{21}}{2}[/tex]Step-by-step explanation:For the equation [tex](x+4)^2-3(x+4)-3=0[/tex] use the substitution [tex]u=x+4.[/tex] Then the equation will take look[tex]u^2-3u-3=0.[/tex]Solve this quadratic equation:[tex]D=(-3)^2-4\cdot 1\cdot (-3)=9+12=21,\\ \\u_{1,2}=\dfrac{-(-3)\pm \sqrt{21}}{2}=\dfrac{3\pm\sqrt{21}}{2}.[/tex]Thus, [tex]x+4=\dfrac{3-\sqrt{21}}{2}\text{ or }x+4=\dfrac{3+\sqrt{21}}{2},\\ \\x_1=\dfrac{3-\sqrt{21}}{2}-4=\dfrac{-5-\sqrt{21}}{2}\text{ or }x_2=\dfrac{3+\sqrt{21}}{2}-4=\dfrac{-5+\sqrt{21}}{2}.[/tex]