Question

1. Find the value of k, for which one root of the quadratic equation kx2-14x+8=0 is
six times the other.
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Answer 1

Answer:

Let roots be a and B

A/q

a = 6B

now, if a and Bare roots then equation will be (x -a)(x -B) =0

(x -a)(x -B) =0

.2 → x² - (a+B)x + aß =0

now putting a = 6B,

= x² - (6B +B)x + 6B×B =0

=x* - 7Bx +6B² =0

now comparing with kx2-14x +8 =0

7B =14/k

→B =2/k

→B? = 4/k?_ and 6B2 =8/k

(1)

→B² =4/3k_ (2)

equating (1) and (2), we get,

4/k? = 4/3k

k=3