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Questions
The following numbers, K + 3, K + 2, 3K – 7 and 2K – 3 are in proportion. Find k.
The numbers k + 3, k + 2, 3k – 7 and 2k – 3 are in proportion find k.
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Solution
k + 3, k + 2, 3k − 7 and 2k − 3 are in proportional then
⇒ `(k + 3)/(k + 2) = (3k - 7)/(2k - 3)`
⇒ (k + 3) (2k − 3) = (k + 2) (3k − 7)
Left side:
(k + 3) (2k − 3) = 2k2 − 3k + 6k − 9
Right side:
(k + 2) (3k − 7) = 3k2 − 7k + 6k − 14
⇒ 2k2 − 3k + 6k − 9 = 3k2 − 7k + 6k − 14
⇒ 2k2 + 3k − 9 = 3k2 − k − 14
⇒ 3k2 − k − 14 − 2k2 − 3k + 9 = 0
⇒ k2 − 4k − 5 = 0
⇒ k2 − 5k + k − 5 = 0
⇒ k (k − 5) + 1(k − 5) = 0
⇒ k = 5 or k = −1
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