Question

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve

Answer 1

Answer:

$x=\frac{-7+\sqrt{15}}{2}$

$x=\frac{-7-\sqrt{15}}{2}$

Step-by-step explanation:    

we have

$(2x+3)^2+8(2x+3)+1=0$

Let

$u=(2x+3)$

substitute the variable

so

$u^2+8u+1=0$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$u^2+8u+1=0$

so

$a=1\\b=8\\c=1$

substitute in the formula

$u=\frac{-8(+/-)\sqrt{8^{2}-4(1)(1)}} {2(1)}$

$u=\frac{-8(+/-)\sqrt{60}} {2}$

$u=\frac{-8(+/-)2\sqrt{15}} {2}$

$u=-4(+/-)\sqrt{15}$

so

$u_1=-4+\sqrt{15}$

$u_2=-4-\sqrt{15}$

Find the value of x

Remember that

$u=(2x+3)$

First solution

$-4+\sqrt{15}=(2x+3)$

$2x=-4+\sqrt{15}-3$

$2x=-7+\sqrt{15}$

$x=\frac{-7+\sqrt{15}}{2}$

Second solution

$-4-\sqrt{15}=(2x+3)$

$2x=-4-\sqrt{15}-3$

$2x=-7-\sqrt{15}$                

$x=\frac{-7-\sqrt{15}}{2}$