Find the value of K if one of the root of quadratic eqn Kx2-14x+8=0 is six times to the other?

Find the value of K if one of the root of quadratic eqn Kx2-14x+8=0 is

1.	Find the value of K if one of the root of quadratic eqn Kx2-14x+8=0 is six times to the other?
Answer» Kx2\xa0- 14x + 8 = 0\xa0On Comparing with standard form of polynomial, we get a = k, b = -14 and c = 8\xa0let one root =\xa0{tex}\\alpha {/tex}then another root =\xa0{tex}6\\alpha {/tex}We know,Sum of roots =\xa0{tex}{-b\\over a}{/tex}=>\xa0{tex}\\alpha + 6\\alpha = {14\\over K} \\\\=> 7\\alpha = {14\\over K}\\\\=> \\alpha = {2\\over K} \\ ... (i){/tex}Also,\xa0Product of roots =\xa0{tex}c\\over a{/tex}{tex}\\alpha \\times 6\\alpha = {8\\over K}\\\\6\\alpha ^2 = {8\\over K}{/tex}{tex}6\\times ({2\\over K})^2= {8\\over K} \\ using \\ (i) \\\\=> {24\\over K^2} = {8\\over K} \\\\=> K = 3 {/tex}