The solution to the original equation will be -2 and -6
Given the function [tex](x + 3)^2 + (x + 3) - 2 = 0.[/tex]
If u = x + 3, the function in terms of "u" will be;
[tex]u^2+u-2=0[/tex]
Factorize the resulting equation as shown:
[tex]u^2+u-2=0\\u^2+2u-u-2=0\\u(u+2)-1(u+2)=0\\(u-1)(u+3) = 0\\u=1 \ and \ -3[/tex]
Since x + 3 = u
x = u - 3
If u = 1
x = 1 - 3 = -2
If u = -3
x = -3 - 3 = -6
Hence the solution to the original equation will be -2 and -6
Correct 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?
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