# Rational numbers?

• Nov 25th 2010, 03:43 AM
IoNForce
Rational numbers?
So yesterday, on math class, I solved a difficult task (a quadratic polynomial). I solved it correctly, but at one point teacher asked me something I couldn't explain. The task says: "The biggest value of the quadratic polynomial is 3/2 and one of his zeros is √3 - 1. Determine the polynomial if it's coefficients are rational numbers!". The first zero was √3 - 1, and I knew the other one had to be -√3 - 1 (I know it had to be it's conjugate). Teacher said that these are NOT complex numbers - so i cant conjugate anything. How can I prove that the rational coefficient [√3 - 1] is equal(?) to the rational coefficient [-√3 - 1]?
• Nov 25th 2010, 05:04 AM
earboth
Quote:

Originally Posted by IoNForce
So yesterday, on math class, I solved a difficult task (a quadratic polynomial). I solved it correctly, but at one point teacher asked me something I couldn't explain. The task says: "The biggest value of the quadratic polynomial is 3/2 and one of his zeros is √3 - 1. Determine the polynomial if it's coefficients are rational numbers!". The first zero was √3 - 1, and I knew the other one had to be -√3 - 1 (I know it had to be it's conjugate). Teacher said that these are NOT complex numbers - so i cant conjugate anything. How can I prove that the rational coefficient [√3 - 1] is equal(?) to the rational coefficient [-√3 - 1]?

1. The quadratic equation $y = ax^2+bx+c$ can be rearranged to $y = a(x-x_v)^2+y_v$ where $V(x_v, y_v)$ is the vertex of the corresponding parabola.

2. For symmetry's reason you know that the mean of the zeros equals $x_v$. From $\sqrt{3}-1$ you can get two different possible zeros: $\sqrt{3}+1$ or $-\sqrt{3}-1$

Calculating the mean value you'll get:

$\dfrac{(\sqrt{3}-1)+(\sqrt{3}+1)}2 = \sqrt{3}$ which is definitely no rational number;
or
$\dfrac{(\sqrt{3}-1)+(-\sqrt{3}-1)}2 = -1$ which is a valid number for $x_v$ according to the text of your question.

3. The maximum value occurs at the vertex. Therefore you know now:

$y = a(x+1)^2+\frac32$

4. Plug in the value of the known zero and solve for a:

$0=a(\sqrt{3}-1+1)^2+\frac32$

I've got $a = -\frac12$
• Nov 25th 2010, 05:11 AM
IoNForce
Quote:

Originally Posted by earboth
1. The quadratic equation $y = ax^2+bx+c$ can be rearranged to $y = a(x-x_v)^2+y_v$ where $V(x_v, y_v)$ is the vertex of the corresponding parabola.

2. For symmetry's reason you know that the mean of the zeros equals $x_v$. From $\sqrt{3}-1$ you can get two different possible zeros: $\sqrt{3}+1$ or $-\sqrt{3}-1$

Calculating the mean value you'll get:

$\dfrac{(\sqrt{3}-1)+(\sqrt{3}+1)}2 = \sqrt{3}$ which is definitely no rational number;
or
$\dfrac{(\sqrt{3}-1)+(-\sqrt{3}-1)}2 = -1$ which is a valid number for $x_v$ according to the text of your question.

3. The maximum value occurs at the vertex. Therefore you know now:

$y = a(x+1)^2+\frac32$

4. Plug in the value of the known zero and solve for a:

$0=a(\sqrt{3}-1+1)^2+\frac32$

I've got $a = -\frac12$

Yeah I've got $a = -\frac12$ too, but I have to prove the bold text. The reason why can I get the other two possible zeros from $\sqrt{3}-1$.
• Nov 25th 2010, 05:33 AM
Soroban
Hello, IoNForce!

Quote:

So yesterday, on math class, I solved a difficult task (a quadratic polynomial).
I solved it correctly, but at one point teacher asked me something I couldn't explain.

The task says: "The biggest value of the quadratic polynomial is 3/2
and one of its zeros is √3 - 1.
Determine the polynomial if it's coefficients are rational numbers."

The first zero was √3 - 1, and I knew the other one had to be -√3 - 1.
(I know it had to be its conjugate).
Teacher said that these are NOT complex numbers - so i can't conjugate anything.

You are right . . . Irrational nuimbers have conjugates, too.

This is the way I approached it . . .

The two roots are: . $\text{-}1 + \sqrt{3}\,\text{ and }\,\text{-}1 - \sqrt{3}$

A quadratic polynomial with these two root has the form:

. . $f(x) \;=\;a\bigg[x - (-1 + \sqrt{3})\bigg]\,\bigg[x -(-1 - \sqrt{3})\bigg]$

. . . . . . $=\;a\bigg[(x-1) - \sqrt{3}\bigg]\,\bigg[(x-1) + \sqrt{3})\bigg]$

. . . . . . $=\;a\bigg[(x-1)^2 - (\sqrt{3})^2\bigg]$

. . . . . . $=\;a(x^2 - 2x + 1 - 3) \;=\;a(x^2 - 2x - 2)$

We have: . $f(x) \;=\;ax^2 - 2ax - 2a$

The maximum value $(\frac{3}{2})$ occurs at its vertex.

The vertex is at: . $x \:=\:\frac{-(\text{-}2a)}{2(a)} \:=\:1$

. . and: . $f(1) \:=\:a(1^2) - 2a(1) - 2a \:=\:\text{-}3a$

Since the maximum vaue is $\frac{3}{2}$, we have: . $\text{-}3a \:=\:\frac{3}{2} \quad\Rightarrow\quad a \:=\:\text{-}\frac{1}{2}$

Therefore: . $f(x) \;=\;\text{-}\frac{1}{2}(x^2 + 2x - 2) \;=\;\text{-}\frac{1}{2}x^2 - x + 1$

• Nov 25th 2010, 05:47 AM
Plato
The two roots are $-1+\sqrt{3}~\&~-1-\sqrt{3}$.
So the equation is $x^2-[(-1+\sqrt{3})+(-1-\sqrt{3})]x+(-1+\sqrt{3})(-1-\sqrt{3})=0$.