Find the values of k for which the quadratic equation 9x^2 - kx + (k - 7) = 0 has
i) one root is twice the value of the other.
How should i solve from here? Should i use products of roots or sum of roots?
Thanks.
You could use the sum of the roots,
since if one root is twice the other,
then the sum of the roots is 3 times the smaller root.
$\displaystyle r_1+r_2=r_1+2r_1=3r_1$
then write the factors using these roots written in terms of k,
finally use the product of the roots to write the values of k.
That's one way.
Let a be one of the roots and 2a the other. Then we must have $\displaystyle 9x^2 - kx + (k - 7) = 9(x- a)(x- 2a)= 9x^2- 27ax+ 18a^2$.
That is, k= 27a and $\displaystyle k- 7=18a^2$. Since k= 27a, $\displaystyle 27a- 7= 18a^2$ or $\displaystyle 18a^2- 27a+ 7= 0$.