Monoxdifly
MHB
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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?
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?