Find the value of p, for which one root of the quadratic equation px^2-14x+8=0 is 6 times the other
Answers
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.
let us take the roots to be "a" and "6a"
the quadratic equation is : px2 - 14x+8 = 0
we know that the sum of zeroes is -b/a
a+6a = 14/p
7a = 14/p
a = 2/p eq.1
product of roots = c/a
6a2 = 8/p ( putting the value of a from eq.1)
6(2/p)2 = 8/p
6*4/p2 = 8/p
24/p2 = 8/p
one p gets cancelled :
p = 24/8 that is 3
therefore the value of p is 3
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