Question

Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u – 2 = 0 u2 + u + 1 = 0 Factor the equation. What are the solutions of the original equation?

Answer 1

Answer: x= -5 , x= -2

Answer 2

The solutions of the original equation are x=-5 and x=-2

Explanation:

we have

`(x+3)^2+(x+3)-2=0`

Let

`u=(x+3)`

Rewrite the equation

`(u)^2+(u)-2=0`

Complete the square

`u^2+u=2`

`u^2+u+1/4=2+1/4`

`u^2+u+1/4=9/4`

rewrite as perfect squares

`(u+1/2)^2=9/4`

square root both sides

`(u+1/2)=\pm(3)/(2)`

`u=(-1/2)\pm(3)/(2)`

`u=(-1/2)+(3)/(2)=1`

`u=(-1/2)-(3)/(2)=-2`

the solutions are

u=-2,u=1

Alternative Method

The formula to solve a quadratic equation of the form

`ax^(2) +bx+c=0`

is equal to

`x=\frac{-b\pm\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`(u)^2+(u)-2=0`

so

`a=1\\b=1\\c=-2`

substitute in the formula

`u=\frac{-1\pm\sqrt{1^(2)-4(1)(-2)}} {2(1)}`

`u=\frac{-1\pm√(9)} {2}`

`u=\frac{-1\pm3} {2}`

`u=\frac{-1+3} {2}=1`

`u=\frac{-1-3} {2}=-2`

the solutions are

u=-2,u=1

Find the solutions of the original equation

For u=-2

`-2=(x+3)` ---->
`x=-2-3=-5`

For u=1

`1=(x+3)` ---->
`x=1-3=-2`

therefore

The solutions of the original equation are

x=-5 and x=-2