How to think while solving question like this? If $ax^2+bx+c=0$ has roots $A$ and $A^n$, then $(ac^n)^{1/(n+1)}+(a^nc)^{1/(n+1)}+b=0$.

Question

I want to know what should be your thinking while solving this question.

$ax^2 + bx + c = 0$ has roots $A$ and $A^n$. Prove that $$(ac^n)^{1/(n+1)} + (a^nc)^{1/(n+1)} +b = 0$$

What are you thinking in your mind about the variables, about the steps?

You see, we don’t know $n$, $a$, $b$, $c$, neither $A$. But we know a equation and it has to be $0$.

I get these questions in mind:

it that we require to know value of $n$ since that can tells us a huge guess to what are values of roots?
So , in my way of thinking. I feel we need to know $a$, $c$, and $b$ values.

Please do share yours. I do not want a solution. Just how would you think in these types of questions.

My first step is to ask myself, have I ever seen an equation like $ax^2+bx+c=0$, and to refresh my memory about its roots. Lucky for me, I do remember seeing this equation before... — kimchi lover, Apr 11 '21 at 14:28
The immediate line of attack: product of roots is $A^{n+1}=c/a$, so $(a^nc)^{1/n+1}=(a^{n+1}(c/a))^{\frac1n+1}=\dots$ — user10354138, Apr 11 '21 at 14:31
Then I look at the rest of the problem statement and try to connect the new details with what I already know. — kimchi lover, Apr 11 '21 at 14:32
A^n can only be equal to A. If let us say n is greater than 1 , Only if A = 1 is what I think@kimchilover — Rider, Apr 11 '21 at 14:34
@Rider Because pinning it down to a nasty expression isn't going to work very nicely with your exponents. — user10354138, Apr 11 '21 at 14:35
Hmm k makes sense. We don’t need the original value of any variable like that. — Rider, Apr 11 '21 at 14:36
To be clear: Are those exponents $\frac{1}{n}+1$, or $\frac{1}{n+1}$? — Blue, Apr 11 '21 at 14:42

score 3 · Accepted Answer · answered Apr 11 '21 at 15:31

$ax^2 + bx + c = 0\;$ has roots $A$ and $A^n$. Prove that $$\left(ac^n\right)^{\frac1{n+1}}+\left(a^nc\right)^{\frac1{n+1}}+b=0$$

Since $\;\dfrac ca=A\cdot A^n=A^{n+1}\;,\;$ we get that

$\displaystyle\left(ac^n\right)^{\frac1{n+1}}+\left(a^nc\right)^{\frac1{n+1}}+b=$

$=\displaystyle\sqrt[n+1]{ac^n}+\sqrt[n+1]{a^nc}+b=$

$=c\;\sqrt[n+1]{\dfrac ac}+a\;\sqrt[n+1]{\dfrac ca}+b=$

$=c\;\sqrt[n+1]{\dfrac1{A^{n+1}}}+a\;\sqrt[n+1]{A^{n+1}}+b=$

$=\dfrac cA+aA+b=$

$=\dfrac1A\left(c+aA^2+bA\right)=$

$=\dfrac1A\left(aA^2+bA+c\right)=$

$=0\;\;$ indeed $\;A\;$ is a root of $\;ax^2+bx+c\;.$

So we have proved that

$\displaystyle\left(ac^n\right)^{\frac1{n+1}}+\left(a^nc\right)^{\frac1{n+1}}+b=0\;.$

score 1 · Answer 2 · answered Apr 14 '21 at 06:36

(While I often like to use the Viete relations, they aren't strictly necessary here...)

Since we are given the roots of the quadratic equation, we can write $$ a \ · \ (x - A) \ · (x - A^n) \ \ = \ \ ax^2 + bx + c \ \ . $$

This tells us that $ \ b \ = \ a \ · (-A \ - \ A^n) \ \Rightarrow \ a · (A \ + \ A^n) \ + \ b \ = \ 0 \ \ $ and $ \ a· A · A^n \ = \ a·A^{n+1} \ = \ c \ \ . $

Carry on...

score 0 · Answer 3 · edited Apr 11 '21 at 16:14

0

Here is my hand written proof:

edited Apr 11 '21 at 16:14

Angelo

12,328

answered Apr 11 '21 at 15:09

RNR

19

Common why a down vote when I am right. – RNR Apr 11 '21 at 15:27
My friend , I totally upvote it. Thanks. I didn’t downvote – Rider Apr 11 '21 at 15:37
@Rider Thanks man. I appreciate that. – RNR Apr 11 '21 at 15:38
4

Please take the effort to put it into MathJax.
One point of it it's getting searchable ... which is advantageous for worthwhile contributions $\ddot\smile$
– Hanno Apr 12 '21 at 14:21

How to think while solving question like this? If $ax^2+bx+c=0$ has roots $A$ and $A^n$, then $(ac^n)^{1/(n+1)}+(a^nc)^{1/(n+1)}+b=0$.

3 Answers3