Question

if one root of quadratic equation kx2-14x+8 is six times the other then what is the value of k​

Answer 1

Answer:

$\bold\red{k=2}$

Step-by-step explanation:

Given,

a quadratic equation,

$k {x}^{2} - 14x + 8 = 0$

Let the first root be 'm'

Then ,

the other root will be '6m'

Now,

we know that,

sum of roots = -b/a

=> m + 6m = -(-14)/k

=> 7m = 14/k

=> 1/k = m/2 ..........(i)

Also,

product of roots = c/a

=> m × 6m = 8/k

=> 6${m}^{2}$ = 8/k ..........(ii)

Now,

putting the value of 1/k from (i) in (ii),

we get,

$= > 6 {m}^{2} = \frac{8m}{2} = 4m \\ \\ = > 6 {m}^{2} - 4m = 0 \\ \\ = > m(3m - 2) = 0 \\ \\ = > m = 0 \\ \\ and \\ \\ = > m = \frac{2}{3}$

Therefore,

1/k = m/2 = 0/2 = 0

=> k = 1/0 = ∞

It can't be possible.

Therefore,

1/k = m/2 = (2/3)/2 = 1/2

Hence,

k = 2