Solving a Quadratic Equation Problem

[mcsun]

Homework Statement

Solve the equation ax² + bx + c = 0. The following are the given information:
(a) a = 1
(b) One root is twice the other root.
(c) c = 2

You are NOT required to solve the problem but to decide whether:
1) The given information is all needed to solve the problem. In this case write "A" as your solution.
2) The total amount of information is insufficient to solve the problem. If so, write "I" as your solution.
3) The problem can be solved without using one or more of the given pieces of information. In this case, write the letter(s) corresponding to the items not needed.

Solution provided by the textbook:

Answer (I)

Homework Equations

General Quadratic Equation:

ax² + bx + c = 0


Let \alpha and \beta be the roots of the equation ax² + bx + c = 0

Sum of Roots = (\alpha + \beta) = - [\frac{b}{a}]

Product of Roots = (\alpha\beta) = \frac{c}{a}

The Attempt at a Solution

My answer is (A).

Why? I shall poof:

The equation: x² + bx + 2 = 0

Firstly determine the value of "b":

x² + bx + 2 = 0
[ax² + bx + c = 0]

Sum of its Roots = (\alpha + \beta) = - [\frac{b}{a}] = - [\frac{b}{1}] = - b

Product of its Roots = (\alpha\beta) = \frac{c}{a} = \frac{2}{1} = 2

Since the question states that "One root is twice the other", therefore...

\beta = 2\alpha

Hence,

Sum of its Roots = (\alpha + 2\alpha) = 3(\alpha) = -b

Product of its Roots = (\alpha)(2\alpha) = 2(\alpha)² = 2

Since,

2(\alpha)² = 2

\alpha = 1 ---> (1)

3(\alpha) = -b ---> (2)

Substitute (1) into (2):

b = -3

Now substitute [b = -3] into the original quadratic equation:

x² - 3x + 2 = 0
(x - 2)(x - 1) = 0

Hence, x = 2 of x = 1

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Question:
Is the above answer & proof of mine correct? Or am I absolutely wrong?

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TAKE NOTE:
3(\alpha) is NOT EQUAL to 3^(\alpha)

 

[Gib Z]

The textbook is correct - you are not given enough information to uniquely determine the numerical value of the solutions. Look a bit more closely in your working :

2(\alpha)² = 2

\alpha = 1 ---> (1)

Do you see something you might not have accounted for?

 

[mcsun]

Ohhh... I get it now!
Since the question didn't mention the type of Roots that the equation possesses [Real Distinct Roots, Equal Roots, Unreal Roots], I won't be able to solve Line 5.

Line 1: 2(\alpha)² = 2

Line 2: (\alpha)² = 1

Line 3: (\alpha)² - 1 = 0

Line 4: (\alpha + 1)(\alpha - 1) = 0

Line 5: \alpha = 1 OR \alpha = -1


Thanks for the guidance, Gib Z.

 

[Zenster]

Lol i just joined like 5 months late of this thread. Anyways.

Since the question didn't mention the type of Roots that the equation possesses [Real Distinct Roots, Equal Roots, Unreal Roots], I won't be able to solve Line 5.

Line 1: 2()2 = 2

Line 2: ()2 = 1

Line 3: ()2 - 1 = 0

Line 4: ( + 1)( - 1) = 0

Line 5: = 1 OR = -1

First of all these roots are definitely real and distinct since you already solved the possibilities using statement (b). Anyways, you won't be able to solve line 5 even if the question did mention what type of roots they are. Am I wrong? Btw what continent does this question come from? I know the country I just moved to does not touch on alpha beta roots, which is sad.

 

[Gib Z]

Welcome to PF Zenster!

It may be prudent to leave these old threads be =] I'd be interesting to find out what country you live in ! Surely they must just cover it in a class you didn't choose to take, or at a level you haven't learned up to yet. I would assume the original poster was American.

 

[Zenster]

Thanks for the warm welcome Gib Z and the fast response. Well, maybe I have not taken a class that has taught this yet. And this leads into what I have been thinking before. The country (which uses just one education system other than IB) which I came from in Asia, taught alpha beta roots in grade 9 (and some calculus). I moved to Canada few months ago, took grade 11 math now I am currently taking grade 12 math, and I will take calculus next year so its like relearning it for me. So just maybe and hopefully this alpha and beta roots will be taught. I was just interested to know what people were learning here from grade 7 to 10 if they only start on these specific topics in grade 11 and 12. I guess the learning here is more whole rounded and offers a more complete understanding of each chapter, but surely they could not be learning about "algebra" and multiplying/dividing from grade 7 to 10, could they?

I guess I should end my comment here, typed a bit too much at once. Anyways thanks again for the reply. I could really get used to this forum thing.