find the value of K for which one root of the equation kx square - 14 x + 8 = 0 is 6 times of the other
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Solution :
Given quadratic equation kx²-14x+8=0
Compare above equation with
ax²+bx+c=0 we get,
a = k , b = -14 , c = 8
Let One root = m
Second root = 6m
i ) Sum of the tee 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 = 8/k
=> 6m² = 8/k
=> 6( 2/k )² = 8/k [ From ( 1 ) ]
=> ( 6 ×4 )/k²= 8/k
=> 24/8 = k²/k
=> 3 = k
Therefore ,
k = 3
•••••
Given quadratic equation kx²-14x+8=0
Compare above equation with
ax²+bx+c=0 we get,
a = k , b = -14 , c = 8
Let One root = m
Second root = 6m
i ) Sum of the tee 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 = 8/k
=> 6m² = 8/k
=> 6( 2/k )² = 8/k [ From ( 1 ) ]
=> ( 6 ×4 )/k²= 8/k
=> 24/8 = k²/k
=> 3 = k
Therefore ,
k = 3
•••••
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