Question

Find the value of k for which one root of the quadratic equation kx^2-4x+ 8 -0 is six times the other

Answer 1

Answer:

12/49

Step-by-step explanation:

Let the two zeroes be y& 6y.

Sum of the zeroes=y+6y    =7y

Product of the zeroes=y*6y  =6y^2

Sum of the zeroes using coefficients=-b/a = -(-4)/k =4/k

Product of the zeroes using coefficients= c/a =8/k

So,

Sum: 7y =4/k

         y =4/7k

       y^2=16/49k^2.........eq.1    (squaring both sides)

Product:  6y^2 =8/k

               y^2 =8/6k

                y^2=4/3k........eq.2

From eq.1 and eq.2,

           16/49k^2=4/3k

           16/4=49k^2/3k

                4=49k/3

                12/49=k  

Hope this helps...

Answer 2

Answer: k=3

Step-by-step explanation:

Let the first root be α.

then second root will be 6​α.

Sum of roots = −ba             α+6α= −(14)k=14k                   7α=14k                     α=2kandProduct of roots=ca                   α×6α=8k                        6α2=8k                  6(2k)2=8k                            k=3