Q4 of 83 Page 76





⇒ 6(x2 + 1) = 13x


⇒ 6x2 + 6 = 13x


⇒ 6x2 – 13x + 6 = 0


⇒ 6x2 – 9x – 4x + 6 = 0


taking 3x common from first two terms and – 2 common from next two


⇒ 3x(2x – 3) – 2(2x – 3) = 0


⇒ (3x – 2)(2x – 3) = 0


⇒ 3x – 2 = 0 or 2x – 3 = 0


⇒ 3x = 2 or 2x = 3


Therefore and


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