Question
Solve the following quadratic equation:$x^2 + 6x - (a^2 + 2a - 8) = 0$
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Answer
$x^2 + 6x - (a^2 + 2a - 8) = 0$
$\Rightarrow x^2 + 6x - (a^2 + 4a - 2a - 8) = 0$
$\Rightarrow x^2 + 6x - [a(a + 4) - 2(a + 4)] = 0$
$\Rightarrow x^2+ 6x - (a + 4)(a - 2) = 0$
$\Rightarrow x^2 + (a + 4)x - (a - 2)x - (a + 4)(a - 2) = 0$
$\Rightarrow x[x + (a + 4)] - (a - 2)[x + (a + 4)] = 0$
$\Rightarrow [x + (a + 4)][x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0$ or $x - (a - 2) = 0$
$\Rightarrow x = -(a + 4)$ or $x = (a - 2)$
$\Rightarrow x^2 + 6x - (a^2 + 4a - 2a - 8) = 0$
$\Rightarrow x^2 + 6x - [a(a + 4) - 2(a + 4)] = 0$
$\Rightarrow x^2+ 6x - (a + 4)(a - 2) = 0$
$\Rightarrow x^2 + (a + 4)x - (a - 2)x - (a + 4)(a - 2) = 0$
$\Rightarrow x[x + (a + 4)] - (a - 2)[x + (a + 4)] = 0$
$\Rightarrow [x + (a + 4)][x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0$ or $x - (a - 2) = 0$
$\Rightarrow x = -(a + 4)$ or $x = (a - 2)$
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