Question

find the value of k for which one root of the quadratic equation kx square - 14 x + 8 = 20 is 6 times the other .....

plz try to answer this

Answer 1

k=12

how?

p(x)=kx2-14x-2=0

so α+β=-b/a

=14/k(α=6α)

so α+6α=14/k

7α=14/k

α=2/k

αβ=c/a

-2/k

α•6α=-2/k

6α^2=-2/k

3α^2=-1/k

α^2=1/3k

4/k^2=1/3k

4/k=1/3

k=12

Answer 2

Solution :

Given quadratic equation kx²-14x+8=20

Compare above equation with

ax²+bx+c=0 we get,

a = k , b = -14 , c = -12

Let One root = m

Second root = 6m

i ) Sum of the two roots = -b/a

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

=> 7m = 14/k

=> m = 14/7k

=> m = 2/k ---( 1 )

ii ) product of the roots = c/a

=> m × 6m = -12/k

=> 6m² = -12/k

=> 6( 2/k )² = 8/k [ From ( 1 ) ]

=> ( 6 ×4 )/k²= -12/k

=> 24/-12 = k²/k

=> -2 = k

Therefore ,

k = -2