1. ## Inequalities

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.

2. Originally Posted by fantasylo
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.

$r_1+r_2=r_1+2r_1=3r_1$

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.

3. Originally Posted by fantasylo
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.
Use the quadratic formula to find the two roots:

$x=\dfrac{k\pm\sqrt{k^2-36(k-7)}}{18}$

Then assuming the roots are real, you have:

$\dfrac{k+\sqrt{k^2-36(k-7)}}{18}=2\ \dfrac{k - \sqrt{k^2-36(k-7)}}{18}$

after simplifying you are done.

CB

4. Let a be one of the roots and 2a the other. Then we must have $9x^2 - kx + (k - 7) = 9(x- a)(x- 2a)= 9x^2- 27ax+ 18a^2$.

That is, k= 27a and $k- 7=18a^2$. Since k= 27a, $27a- 7= 18a^2$ or $18a^2- 27a+ 7= 0$.

5. Originally Posted by Archie Meade
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.

$r_1+r_2=r_1+2r_1=3r_1$

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.
and remember this....
if $r_1$ and $r_2$ are roots of the equation then $r_1+r_2=-\frac{b}{a}$