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