quadratic equations using factorisation
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Let (x+3) = a
=> a^2 - 4a - 5 = 0
=> a^2 - 5a + a - 5 = 0
=> a ( a - 5) +1 (a-5) = 0
=> (a + 1) (a - 5) = 0
=> ( x + 3 + 1) (x + 3 - 5) = 0
=> ( x + 4) (x - 2) = 0
Roots are 2 and - 4
=> a^2 - 4a - 5 = 0
=> a^2 - 5a + a - 5 = 0
=> a ( a - 5) +1 (a-5) = 0
=> (a + 1) (a - 5) = 0
=> ( x + 3 + 1) (x + 3 - 5) = 0
=> ( x + 4) (x - 2) = 0
Roots are 2 and - 4
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