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5 days ago

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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?

2 answers:

lawyer [4K]5 days ago

6 0

Answer:

Part 1 gives x + 3

Part 2 yields B u2 + u - 2 + 0

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

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

Step-by-step explanation:

on edg... Good Luck!!!

AnnZ [3.8K]5 days ago

4 0

Answer:

The roots of the initial equation are x=-5 and x=-2

Step-by-step explanation:

we have

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

Let

$u=(x+3)$

Rephrase 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 complete squares

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

take the square root of 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

Alternative Method

The equation used to solve a quadratic of the type

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

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this case we have

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

so

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

insert into 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 discovered roots are

u=-2,u=1

Find the roots of the initial equation

For u=-2

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

For u=1

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

therefore

The roots of the initial equation are

x=-5 and x=-2

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