Q27 of 47 Page 1

Solve for x:

Given,

⇒


⇒ 3 (x2 – 11x + 28 + x2 – 11x + 30) = 10 (x2 – 12x + 35)


⇒ 3 (2x2 – 22x + 58) = 10x2 – 120x + 350


⇒ 6x2 – 66x + 174 = 10x2 – 120x + 350


⇒ 4x2 – 54x + 176 = 0


⇒ 2x2 – 27x + 88 = 0


[By splitting the middle term]


⇒ 2x2 – 16x – 11x + 88 = 0


⇒ 2x (x – 8) – 11 (x – 8) = 0


⇒ (x – 8) (2x – 11) = 0


⇒ x – 8 = 0 or 2x – 11 = 0


⸫ x = 8 or x =


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