sec 3 amath question

smileyrainbow SG

its on the quadratic equation topic.

The equation 18e² + e^2x = 9e^1+xhas roots α and β.

Show that e ^{α- β} = 2

my first step is

(e^x)² – 9e (e^x) + 18e² = 0

then how to continue?

* those unbold numbers are superscripts.

Mikethm SG

If you solve the equation for x. you will get...

x= ln (6e) or ln (3e)

therefore the roots are ln(6e) or ln(3e)

Scenario No1. e^[ln(6e)-ln(3e)] = e^ln2 = 2 ( as required)

Scenario No.2 e^[ln(3e)-ln(6e)] = 1/(e^ln2) = 1/2 ( not as required)

My question is... why can't scenario No. 2 be true as implied by the question?

Anyone can tell me the flaw in my reasoning?

eagle SG

I think it just means that alpha must be ln (6e) and beta must be ln (3e)

and the other way around just cannot hold true.

Mikethm SG

I think it just means that alpha must be ln (6e) and beta must be ln (3e)

and the other way around just cannot hold true.

But there is nothing that says alpha must be ln(6e). That's what confusing me. I am wondering why alpha cannot be ln(3e)...

wee_ws SG

Hi,

The question has an ambiguity, because e^(alpha - beta) can take either 2 or 1/2, depending on what alpha and beta are.

The setter may not be aware, or he/she is looking at one case only.

Thanks!

Cheers,
Wen Shih

eagle SG

agree with mr wee... think i didn't phrase properly

anyway... Perhaps we could change the question to include one more condition: alpha > beta

crimson soldier SG

Just a simple question what.....

Many questions in the O lvels and A levels are like that. just assume that alpha> beta loh...won't get u anyway if you dispute their "correctness"

Mikethm SG

Just a simple question what.....

Many questions in the O lvels and A levels are like that. just assume that alpha> beta loh...won't get u anyway if you dispute their "correctness"

I have never seen an actual O level Additional Mathematics question which is not 100% mathematically correct.

In fact, I always tell my students that if the question say "take off a shoe and wave it in the air."... please do not take off both shoes and wave them in the air.

wee_ws SG

Hi,

Alpha and beta are arbitrary roots, and (alpha + beta) and (alpha x beta) can be uniquely determined (regardless of alpha > beta, alpha = beta or alpha < beta).

In the question, we need to consider alpha - beta and this cannot be unique. Thus, it seems that the setter is looking at one possible case only.

Anyhow, I'd like to suggest that smileyrainbow highlight the other case [ i.e. e^(alpha - beta) = 1/2 ] to the teacher.

It is a good question that provokes deep thinking nevertheless :)

Thanks!

Cheers,
Wen Shih

Mikethm SG

To add on to what wen shih said in his last post...

As I understand it, the alpha beta type of questions were never meant to be solved by finding the actual roots of the equation. I did learn alpha beta questions in the late late 80s but it wasn't tested in the syllabuses which I took... so I didn't explore it in depth until last year when it came back into syllabus.

This mean that we are supposed to make use of the values of (alpha+beta) and (alpha x beta) and never meant to calculate the (possible) values of alpha and beta.

I was drawn to this question because I could not think of a way to calculate (alpha-beta) without 1st calculating the roots (which in a correctly setted alpha beta question would not be the case).

My assumption was that this question is correct and needed some other concepts which I failed to use. Thus my interest as I want to plug any gap in my thinking processes. But the evidence so far by consesus is that the question is flawed.

eagle SG

it's possible that

1) TS didn't post the full question out. Happens very commonly around here. Some of new forumers who didn't read the homework forum rules I posted only post out a part of their question, expecting everyone to understand fully.

2) The question is indeed set inaccurately.

We need TS to confirm where his question comes from.

I did learn alpha beta questions in the late late 80s but it wasn't tested in the syllabuses which I took

How old are you???

smileyrainbow SG

This mean that we are supposed to make use of the values of (alpha+beta) and (alpha x beta) and never meant to calculate the (possible) values of alpha and beta.

I was drawn to this question because I could not think of a way to calculate (alpha-beta) without 1st calculating the roots (which in a correctly setted alpha beta question would not be the case).

Hi thanks everyone for your replies(:

yeah we are supposed to make use of the values of (alpha+beta) and (alpha x beta) as what mikethm said.. but i have no idea how to do it. because im not sure how to go about finding these values.

eagle SG

you must note that your alpha and beta refers to the roots for x

But if you are to use the formulas given to you for sum of roots and product of roots, it will be the value of e^x instead of x, i.e. your 'roots' of equations will be for e^x

wee_ws SG

Hi,

The usual approaches of finding the sum and product of roots may not work to help us to show that e^(alpha - beta) = 2. Thus, there may be a need to find, by factorisation, the values of e^x which are 6e and 3e. Thanks!

Cheers,
Wen Shih

Lee012lee SG

Question

The equation 18e² + e^2x = 9e^1+xhas roots α and β.

Show that e ^{α- β} = 2 if a > b

Solution

Method 1

18e² + e^2x = 9e^1+x

(e^x)² – 9e (e^x) + 18e² = 0

(e^x – 6e)(e^x – 3e) = 0

e^x = 6e or e^x = 3e

So, x = ln 6e , x = ln 3e

ie a = ln 6e , b = ln 3e

since a > b

Substitute it into e ^{α- β}

e ^α-
β = e^{ln 6 – ln 3}

= e^ln
2

= 2 (Shown)

Method 2

18e² + e^2x = 9e^1+x

(e^x)² – 9e (e^x) + 18e² = 0

Let y = e^X

Let a₁ and b₁ be the roots of y.

Hence,

y² – 9ey + 18e² = 0

Sum of roots = a₁ + b₁ = - b/a = - (9e)/1 = 9e

Product of roots = a₁ × b₁ = c/a = 18e²/1 = 18e²

So, b₁ = 18e² / a₁

Substitute b₁ = 18e² / a₁ into a₁ + b₁ = 9e

a₁ + 18e² / a₁ = 9e

(a₁)² + 18e² = 9ea₁

(a₁)² - 9ea₁ + 18e² = 0

(a₁ – 6e) (a₁ – 3e) = 0

a₁ = 6e or a₁ = 3e

Substitute it into b₁ = 18e² / a₁

b₁ = 3e or b₁ = 6e

Hence the roots of y are 6e and 3e.

Since y = e^X

6e = e^X , 3e = e^X

So, x = ln 6e , x = ln 3e

ie a = ln 6e , b = ln 3e

since a > b

Substitute it into e ^{α- β}

e ^α-
β = e^{ln 6 – ln 3}

= e^ln
2

= 2 (Shown)

Note :

a and b are just the values of x for the given quadratic equation.

So, method 1 finds the values of e^x first by factorisation and

then find the values of x which are a and b.

While method 2 finds the value of y first ie a₁ and b₁ by making

use of the sum of roots and product of roots and substitution method

and then find the values of x which are a and b.

Finally, the values of x in the form of a and b are substituted into the

RHS of e ^{α- β} to get 2 and hence RHS = LHS (Shown).

The topic of alpha and beta in quadratic equation is to study the

relationships among the roots and the two quadratic equations (which

can be of different forms) eg how the roots of a quadratic equation that

can be changed to form another quadratic equation or vice versa and

so on.