Question

Find value of d for which both roots of eq.X^2-6dx+(2-2d+9d^2)=0exceed3

Answer 1

Answer:

This will hold for d > 11 / 9

Step-by-step explanation:

Call the roots a and b, with a being the smaller one.

a + b = - (coeff of x) / (coeff of x²) = 6d, so a, b > 3 tells that we must have d > 1.

Using formula for solutions of quadratic, smaller root is

a = ( 6d - √ ( (6d)² - 4(2-2d+9d²) ) ) / 2 = 3d - √2√(d-1).

So a > 3

<=> 3d - √2√(d-1) > 3

<=> 3(d-1) > √2√(d-1)

<=> 9(d-1)² > 2(d-1)      [ implication both ways since we know d - 1 > 0 ]

<=> 9(d-1) > 2

<=> d > 2/9  + 1 = 11/9