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

[Monoxdifly]

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

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

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

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