Question
Solve the following quadratic equation:$x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
✓
Answer
$x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + (b^2 - 5b + 4b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + [b(b - 5) + 4(b - 5)] = 0$
$\Rightarrow x^2- (2b - 1)x + (b - 5)(b + 4) = 0$
$\Rightarrow x^2 - (2b - 1)x - (b + 4)x + (b - 5)(b + 4) = 0$
$\Rightarrow x[x - (b - 5)] - (b + 4)[x - (b - 5)] = 0$
$\Rightarrow [x - (b - 5)][x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0 or x - (b + 4) = 0$
$\Rightarrow x = b - 5 or x = b + 4$
$\Rightarrow x^2 - (2b - 1)x + (b^2 - 5b + 4b - 20) = 0$
$\Rightarrow x^2 - (2b - 1)x + [b(b - 5) + 4(b - 5)] = 0$
$\Rightarrow x^2- (2b - 1)x + (b - 5)(b + 4) = 0$
$\Rightarrow x^2 - (2b - 1)x - (b + 4)x + (b - 5)(b + 4) = 0$
$\Rightarrow x[x - (b - 5)] - (b + 4)[x - (b - 5)] = 0$
$\Rightarrow [x - (b - 5)][x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0 or x - (b + 4) = 0$
$\Rightarrow x = b - 5 or x = b + 4$
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