Q28 of 49 Page 1

Solve for x:

Given,

⇒


⇒ x (x – 1 + 2x – 4) = 6 (x2 – 3x + 2)


⇒ x (3x – 5) = 6x2 – 18x + 12


⇒ 3x2 – 5x = 6x2 – 18x + 12


⇒ 3x2 – 13x + 12 = 0


⇒ 3x2– 9x – 4x + 12 = 0


⇒ 3x (x – 3) – 4 (x – 3) = 0


⇒ (x – 3) (3x – 4) = 0


⇒ x – 3 = 0 and 3x – 4 = 0


⸫ x = 3 and x =


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