Without solving the equation show it has 2 rational roots

[Jaco Viljoen]

Homework Statement

Without solving the equation 3x^2-8x-3=0 show it has 2 different rational roots.[/B]

Homework Equations

ax^2+bx+c=0

The Attempt at a Solution

I would appreciate if someone would check my work, and advise if I have done the right or wrong thing? Thank you, Jaco
[/B]
3x^2-8x-3=0
a=3 b=-8 c=-3

a*b=-9+1=-8

3x^2-9x+1x-3=0
3x(x-3)+1(x-3)
3x+1=0
x=1/3

x-3=0
x=3

so x=1/3 and x=3

 

[SteamKing]

Homework Statement

Without solving the equation 3x^2-8x-3=0 show it has 2 different rational roots.[/B]

Homework Equations

ax^2+bx+c=0

The Attempt at a Solution

I would appreciate if someone would check my work, and advise if I have done the right or wrong thing? Thank you, Jaco
[/B]
3x^2-8x-3=0
a=3 b=-8 c=-3

a*b=-9+1=-8

3x^2-9x+1x-3=0
3x(x-3)+1(x-3)
3x+1=0
x=1/3

x-3=0
x=3

so x=1/3 and x=3

Well, you did exactly what the question explicitly told you not to do: solve the equation.

You were not to determine that two different solutions to this equation existed by solving for them, but you were supposed to use the various theorems about the roots of polynomials to determine that a.) both roots were real, and b.) both roots were distinct and rational.

Try a different approach.

I'm also moving this thread to the Pre-calculus HW forum, since no integrals or derivatives are required.

 

[Jaco Viljoen]

Good afternoon Steam King,
Thank you for moving the thread, sorry for posting in the wrong place.

I will try again.
Thank you.
Jaco

 

[HallsofIvy]

As SteamKing said, you solved the equation when the problem specifically said "without solving the equation".

What you can do is note that the discriminant for this equation is "$b^2- 4ac= (-8)^2- 4(3)(-3)= 64+ 36= 100$. That is a "perfect square", $10^2$. Do you see why that guarantees the roots of the equation are distinct and rational?

 

[Jaco Viljoen]

@Hallsoflvy Thank you.

 

[SammyS]

@Hallsoflvy I see what you are doing but I don't see what it means for the problem? Sorry if I seem slow its been many years since I did math except the last while trying to get up to speed..

Do you know the quadratic formula ?

If not, do a web search, e.g. google it.

 

[Jaco Viljoen]

Do you know the quadratic formula ?

If not, do a web search, e.g. google it.

b2−4ac
yes, I do.

 

[SammyS]

b2−4ac
yes, I do.

In that green banner, above where you enter the text for your post, there is an X² symbol which allows you to enter a superscript.

b2−4ac becomes b² − 4ac .

 

[geoffrey159]

Study the variations of $f(x) = 3x^2 - 8x - 3$, it will solve the part 'without solving the equation, show there are exactly two roots'

 

[HallsofIvy]

What do you mean by "variations of f"?

 

[geoffrey159]

Hello,
There is a domain over which f is strictly increasing, and another one where f is strictly decreasing. When the change of variation occurs, f is strictly negative, which means it crosses the x-axis 2 times exactly.

 

[HallsofIvy]

b2−4ac
yes, I do.

Strictly speaking that is NOT "the quadratic formula". The quadratic formula, for solutions of the quadratic equation "$ax^2+ bx+ c= 0$" is "$\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$". The discriminant of that, which is what I referred to, is $b^2- 4ac$, the quantity inside the square root. If that is positive, then it has two real values, if 0, a single value, if negative, two complex roots.

 

[HallsofIvy]

Oh, I see- that would involve "completing the square" which I would consider awfully close to "solving the equation! But, yes, that would work.

 

[geoffrey159]

showing the rationality of the solution is more complicated if you don't have the right to use the discriminant.
You can show that $f(\frac{p}{q}) = 0 \Rightarrow 3|pq \text{ and } 8|(p-q)(p+q)$. By trial and error, you find that $(p,q)\in\{(-1,3),(3,1)\}$ satisfy that condition, and they solve your equation while being distinct. So your two roots are rational.

 

[Jaco Viljoen]

Thank you every one.

 

[Mark44]

showing the rationality of the solution is more complicated if you don't have the right to use the discriminant.
You can show that $f(\frac{p}{q}) = 0 \Rightarrow 3|pq \text{ and } 8|(p-q)(p+q)$. By trial and error, you find that $(p,q)\in\{(-1,3),(3,1)\}$ satisfy that condition, and they solve your equation while being distinct. So your two roots are rational.

I'm not sure that the OP will understand your explanation. The problem can be done solely by the use of the discriminant of the quadratic formula. If the discriminant turns out to be a perfect square or the square of a rational number, that is almost enough to say that the two roots are rational.

 

[geoffrey159]

I'm not sure that the OP will understand your explanation. The problem can be done solely by the use of the discriminant of the quadratic formula. If the discriminant turns out to be a perfect square or the square of a rational number, that is almost enough to say that the two roots are rational.

Hello,
I agree that the discriminant proof is the best one, but I understood "how would you prove that without any knowledge about second degree equations ?"
What is an OP ?

 

[SammyS]

Hello,
I agree that the discriminant proof is the best one, but I understood "how would you prove that without any knowledge about second degree equations ?"
What is an OP ?

Original Post or Original Poster .

 

[Mark44]

Hello,
I agree that the discriminant proof is the best one, but I understood "how would you prove that without any knowledge about second degree equations ?"
What is an OP ?

The problem statement didn't say anything about having no knowledge of 2nd degree equations. My read of what Jaco wrote (quoted below) was that there is an implicit assumption that the poster knows how to solve quadratic equations (either by the use of the Quadratic Formula or factorization), but is supposed to answer the question without using these techniques.

Without solving the equation 3x^2-8x-3=0 show it has 2 different rational roots.

 

[pasmith]

We have $x^2 - \frac83 x - 1 = 0$. That the constant term is -1 tells us that the product of the roots is -1. These consequences follow:

Firstly, the roots are not complex conjugates (the product of complex conjugates is real and nonnegative) so the roots are real.

Secondly, the roots must be distinct (if $x^2 = -1$ then $x$ is not real).

Thirdly, neither root is zero and either both roots are rational or both are irrational ($x \neq 0$ is rational if and only if $-1/x$ is rational).

Setting $y = \frac x3$ yields $9y^2 - 8y - 1 = 0$ to which $y = 1$ is a solution by inspection. Thus 3 is a root of the original quadratic and is rational, and hence so is the other.

I don't know if there's a way to rule out the possibility of irrational roots without finding one of the roots (or calculating the discriminant, which comes very close to actually solving the equation).

 

[pasmith]

We have $x^2 - \frac83 x - 1 = 0$. That the constant term is -1 tells us that the product of the roots is -1. These consequences follow:

Firstly, the roots are not complex conjugates (the product of complex conjugates is real and nonnegative) so the roots are real.

Secondly, the roots must be distinct (if $x^2 = -1$ then $x$ is not real).

Thirdly, neither root is zero and either both roots are rational or both are irrational ($x \neq 0$ is rational if and only if $-1/x$ is rational).

Setting $y = \frac x3$ yields $9y^2 - 8y - 1 = 0$ to which $y = 1$ is a solution by inspection. Thus 3 is a root of the original quadratic and is rational, and hence so is the other.

I don't know if there's a way to rule out the possibility of irrational roots without finding one of the roots (or calculating the discriminant, which comes very close to actually solving the equation).

 

[statdad]

"I don't know if there's a way to rule out the possibility of irrational roots without finding one of the roots"

All of the coefficients are integers. What can you conclude about the roots if $b^2 - 4ac$ is zero or a perfect square?

 

[geoffrey159]

You can prove the solutions are rational just with arithmetics$f(\frac{p}{q}) = 0 \iff p(3p-8q) = 3q^2$
$\text{gcd}(p,q) = 1 \iff \text{gcd}(p,q^2) = 1$
$\text{gcd}(p,q^2) = 1 \text{ and } p | 3 q^2 \Rightarrow p | 3 \Rightarrow p\in\{\pm 1, \pm 3\}$.

Case p = 1:
$\text{gcd}(p,3) = 1 \Rightarrow 3 | (3p - 8q) \Rightarrow 3 | 8q \Rightarrow 3 |q$
Also $q (3q + 8 ) = 3 \Rightarrow q | 3$
$3 | q$ and $q | 3$ implies $q = \pm 3$. But $3q + 8 | 3$ so $q = -3$

Case p = -1: same idea

Case p = 3:
$9 = q (q+8) \Rightarrow q | 9 \Rightarrow q \in \{ \pm 1,\pm 3,\pm 9\}$. Since $q + 8 | 9$, then $q = 1$ or $q = - 9$. But p and q are relatively prime so $q = 1$

Case p = -3: same idea

(3,1) and (1,-3) are distinct solutions, there are 2 real solutions, so they are rational