Solve Quadratic Equation: Find c-a Given p and q

[Monoxdifly]

Given p and q are the roots of the quadratic equation $$ax^2-5x+c=0$$ with $$a\neq0$$. If $$p,q,\frac1{8pq}$$ forms a geometric sequence and $$log_a18+log_ap=1$$, the value of c – a is ...
A. $$\frac13$$
B. $$\frac12$$
C. 3
D. 5
E. 7

Since $$p,q,\frac1{8pq}$$ is a geometric sequence, then:
$$\frac{q}{p}=\frac{\frac1{8pq}}q$$
$$\frac{q}{p}=\frac1{8pq^2}$$
$$q=\frac1{8q^2}$$
$$q^3=\frac18$$
$$q=\frac12$$

Also, since $$log_a18+log_ap=1$$, then:
$$log_a18p=log_aa$$
18p = a
$$p=\frac{a}{18}$$

This is where the real problem starts. No matter how I substitute, either it will cancel out the a's or p's, or becoming a quadratic equation with no real roots. What should I do?

 

[I like Serena]

Additionally we have $a(x-p)(x-q)=ax^2 - a(p+q)x + apq = ax^2-5x+c$.
So $a(p+q)=5$ and $apq = c$.
If we substitute the $p$ and $q$ that we've found, we can find $a$ and an expression for $c$ in $a$, and finally $c-a$.

 

[Monoxdifly]

a(a/18)(1/2) = c
(a^2)/36 = c
a^2 = 36c

a(p + q) = 5
a(a/18 + 1/2) = 5
(a^2)/18 + 1/2 a = 5
36c/18 + 1/2 a = 5
2c + 1/2 a = 5
1/2 a = 5 - 2c
a = 10 - 4c
c - a = c - (10 - 4c) = -3c - 10
Sorry, still stuck.

 

[skeeter]

$a(p+q) = 5 \implies a\left(\dfrac{a}{18}+\dfrac{1}{2}\right) = 5 \implies a^2+9a-90=0 \implies a = 6$ since $a>0$ (why?)

finally, you have $c = apq$

you should be able to finish from here

btw ... $c-(10-4c) \ne -3c-10$

 

[Monoxdifly]

Ah, I see. Thank you very much! :D