Question:
Find the value of k, for which one root of the quadratic equation kx2 -14x 8=0 is six times the other.
Answer:
Let the first root = a
Then the second root = 6a
Given equation is: kx2 -14x + 8 = 0
Now, a + 6a = 14/k
=> 7a = 14/k
=> a = 14/(7 * k)
=> a = 2/k
and a * 6a = 8/k
=> 6a2 = 8/k
=> 6 * (2/k)2 = 8/k
=> (6 * 4)/k2 = 8/k
=> 24/k2 = 8/k
=> 3/k2 = 1/k
=> k2 = 3k
=> k2 - 3k = 0
=> k(k - 3) = 0
=> k = 0, 3
If k = 0, then no quadratice equation is possible.
Hence, the value of k is 3
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