[ASK] c – a

Algebra 2

[ASK] c – a

Postby Monox D. I-Fly » Fri Mar 12, 2021 11:13 pm

Given p and q are the roots of the quadratic equation [tex]ax^2-5x+c=0[/tex] with [tex]a\neq0[/tex]. If [tex]p,q,\frac1{8pq}[/tex] forms a geometric sequence and [tex]log_a18+log_ap=1[/tex], the value of c – a is ....

A. [tex]\frac13[/tex]

B. [tex]\frac12[/tex]

C. 3

D. 5

E. 7


Since [tex]p,q,\frac1{8pq}[/tex] is a geometric sequence, then:

[tex]\frac{q}{p}=\frac{\frac1{8pq}}q[/tex]

[tex]\frac{q}{p}=\frac1{8pq^2}[/tex]

[tex]q=\frac1{8q^2}[/tex]

[tex]q^3=\frac18[/tex]

[tex]q=\frac12[/tex]


Also, since [tex]log_a18+log_ap=1[/tex], then:

[tex]log_a18p=log_aa[/tex]

18p = a

[tex]p=\frac{a}{18}[/tex]


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?
Monox D. I-Fly
 
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Re: [ASK] c – a

Postby Guest » Mon Mar 15, 2021 4:12 am

Since
[tex]p, q, \frac{1}{8pq}[/tex] form a geometric sequence
we have:

[tex]q^2=\frac{p}{8pq}[/tex]
[tex]q=2[/tex]

This is the mistake in your calculations.
Will you be able to solve it?
Guest
 

Re: [ASK] c – a

Postby Monox D. I-Fly » Mon Mar 15, 2021 9:48 pm

How did you get q = 2?

Monox D. I-Fly
 
Posts: 28
Joined: Tue May 22, 2018 1:38 am
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Re: [ASK] c – a

Postby Guest » Tue Mar 16, 2021 3:10 am

If 3 numbers - a,b,c form a geometric sequence then

[tex]b^2=ac[/tex]

In your case [tex]p, q, \frac{1}{8pq}[/tex] form a geometric sequence
So

[tex]q^2 = p\frac{1}{8pq}[/tex]
Guest
 

Re: [ASK] c – a

Postby Guest » Tue Mar 16, 2021 3:12 am

Actually

[tex]q^2=\frac{1}{8q}[/tex]
[tex]q^3=\frac{1}{8}[/tex]
​So [tex]q=\frac{1}{2}[/tex]
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