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Anna11 [10]
3 years ago
14

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?
Mathematics
2 answers:
Artyom0805 [142]3 years ago
6 0

Answer:

Part 1 is x + 3

Part 2 is B u2 + u - 2 + 0

Part 3 is (u + 2)(u - 1) = 0

Part 4 is x = -5 or x = -2

Step-by-step explanation:

on edg... Good Luck!!!

Verdich [7]3 years ago
4 0

Answer:

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

Step-by-step 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\frac{3}{2}

u=(-1/2)\pm\frac{3}{2}

u=(-1/2)+\frac{3}{2}=1

u=(-1/2)-\frac{3}{2}=-2

the solutions are

u=-2,u=1

<em>Alternative Method</em>

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\sqrt{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

<em>Find the solutions of  the original equation</em>

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

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