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Question
If one root of k(x − 1)2 = 5x − 7 is double the other root, show that k = 2 or −25
Solution
The given quadratic equation is
k(x – 1)2 = 5x – 7
k(x2 – 2x + 1) – 5x + 7 = 0
kx2 – 2kx + k – 5x + 7 = 0
kx2 – (2k + 5)x + k + 7 = 0 ......(1)
Let the roots be α and 2α
Sum of the roots α + 2α = `- "b"/"a"`
3α = `-((-(2"k" +5))/"k")`
3α = `(2"k" + 5)/"k"`
α = `(2"k" + 5)/(3"k")` ......(2)
Product of te roots α(2α) = `"c"/"a"`
2α2 = `("k"+ 7)/"k"`
α2 = `("k" + 7)/(2"k")` .....(3)
Using equation (2) and (3) we have
`((2"k" + 5)/(3"k"))^2 = ("k" + 7)/(2"k")`
`(4"k"^2 + 20"k" + 25)/(9"k"^2) = ("k" + 7)/(2"k")`
`(4"k"^2 + 20"k" + 25)/(9"k") = ("k" + 7)/(2)`
2(4k2 + 20k + 25) = 9k(k + 7)
8k2 + 40k + 50 = 9k2 + 63k
9k2 + 63k – 8k2 – 40k – 50 = 0
k2 + 23k – 50 = 0
k2 + 25k – 2k – 50 = 0
k(k + 25) – 2(k + 25) = 0
(k – 2) (k + 25) = 0
k – 2 = 0 or k + 25 = 0
k = 2 or k = – 25
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