Question
Solve equation using factorisation method:
$(x - 3)^2 -4(x +3) -5 = 0$
$(x - 3)^2 -4(x +3) -5 = 0$
✓
Answer
$(x - 3)^2 -4(x +3) -5 = 0$
Let $x + 3 = y$
then $y^2 - 4y - 5 = 0$
$\Rightarrow y^2 - 5y + y - 5 = 0$
$\Rightarrow y(y - 5) + 1 (y -5) = 0$
$\Rightarrow ( y -5) (y +1) = 0$
if $y - 5 = 0$ or $y +1 = 0$
then $y = 5$ or $y =-1$
$⇒ x + 3 = 5$ or $x + 3 =-1$
$⇒ x =2$ or $x = -4$
Let $x + 3 = y$
then $y^2 - 4y - 5 = 0$
$\Rightarrow y^2 - 5y + y - 5 = 0$
$\Rightarrow y(y - 5) + 1 (y -5) = 0$
$\Rightarrow ( y -5) (y +1) = 0$
if $y - 5 = 0$ or $y +1 = 0$
then $y = 5$ or $y =-1$
$⇒ x + 3 = 5$ or $x + 3 =-1$
$⇒ x =2$ or $x = -4$
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