Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other
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Answer :
Given quadratic equation :
px^2 −14x+8=0
Step by step explanation :
Let one of the root of equation be 'a', so other root would be '6a'
Now, Comparing the given equation with ax^2 + bx + c = 0, we get :
a = p
b = 14
c = 8
Sum of the roots = -b/a
a + 6a = 14 / p
7a = 14 / p
Hence, a = 2/p ----- eq. (1)
Now,
Product of the roots = c/a
a × 6a = 8/p
6a^2 = 8/p ------- eq. (2)
Now, Substituting value of a in eq. (2)
6 ( 2/p )^2 = 8/p
6 × 4p^2 = 8/p
Cancelling 'p' on both sides, we get :
24p = 8
Dividing both sides by 8,
p = 3
Thus, Value of p is 3.
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