Question

Find the value of p for which one root of quadratic equation px^2-14x+8=0 is 6 times the other

Answer 1

Answer :

Given quadratic equation :

$\tt{px{}^{2}-14x+8=0}$

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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