Answer:
[tex]x=\frac{-7+\sqrt{15}}{2}[/tex]
[tex]x=\frac{-7-\sqrt{15}}{2}[/tex]
Step-by-step explanation:
we have
[tex](2x+3)^2+8(2x+3)+1=0[/tex]
Let
[tex]u=(2x+3)[/tex]
substitute the variable
so
[tex]u^2+8u+1=0[/tex]
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]u^2+8u+1=0[/tex]
so
[tex]a=1\\b=8\\c=1[/tex]
substitute in the formula
[tex]u=\frac{-8(+/-)\sqrt{8^{2}-4(1)(1)}} {2(1)}[/tex]
[tex]u=\frac{-8(+/-)\sqrt{60}} {2}[/tex]
[tex]u=\frac{-8(+/-)2\sqrt{15}} {2}[/tex]
[tex]u=-4(+/-)\sqrt{15}[/tex]
so
[tex]u_1=-4+\sqrt{15}[/tex]
[tex]u_2=-4-\sqrt{15}[/tex]
Find the value of x
Remember that
[tex]u=(2x+3)[/tex]
First solution
[tex]-4+\sqrt{15}=(2x+3)[/tex]
[tex]2x=-4+\sqrt{15}-3[/tex]
[tex]2x=-7+\sqrt{15}[/tex]
[tex]x=\frac{-7+\sqrt{15}}{2}[/tex]
Second solution
[tex]-4-\sqrt{15}=(2x+3)[/tex]
[tex]2x=-4-\sqrt{15}-3[/tex]
[tex]2x=-7-\sqrt{15}[/tex]
[tex]x=\frac{-7-\sqrt{15}}{2}[/tex]
