Q21 of 47 Page 1

Solve for .

Given,

⇒


⇒ 3 (x2 – 7x + 10 + x2 – 7x + 12) = 10 (x2 – 8x + 15)


⇒ 3 (2x2 – 14x +22) = 10x2 – 80x + 150


⇒ 6x2 – 42x + 66 = 10x2 – 80x +150


⸫ 4x2 – 38x + 84 = 0


⸫ 2x2 – 19x + 42 = 0


[By splitting the middle term]


⇒ 2x2 – 12x – 7x + 42 = 0


⇒ 2x (x – 6) – 7 (x – 6) = 0


⇒ (x – 6) (2x – 7) = 0


⇒ x – 6 = 0 or 2x – 7 = 0


⸫ x = 6 or x =


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