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September 28th, 2019, 06:32 AM   #1
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Quadratic polynomial

Find the quadratic equation which satisfies the condition $\displaystyle x_1 x_2 =x_1 +x_2$.
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September 28th, 2019, 07:34 AM   #2
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I don't understand the problem statement. Quadratic equation in what?

You could write the condition itself as
$\displaystyle x_1 = \frac{x_2}{x_2-1}$
or
$\displaystyle x_2 = \frac{x_1}{x_1-1}$
or
$\displaystyle (x_1-1)(x_2-1)-1 = 0$

Are any of those what you were looking for?
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September 28th, 2019, 08:00 AM   #3
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The equation is : $\displaystyle x^2 -2x+2=0$.
It satisfies the condition $\displaystyle x_1 x_2 = x_1 + x_2$.
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September 28th, 2019, 08:28 AM   #4
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Quote:
Originally Posted by idontknow View Post
The equation is : $\displaystyle x^2 -2x+2=0$.
It satisfies the condition $\displaystyle x_1 x_2 = x_1 + x_2$.
I presume that you have a quadratic equation of the form $\displaystyle y = ax^2 + bx + c$ that has zeros at $\displaystyle x_1$ and $\displaystyle x_2$ such that $\displaystyle x_1 x_2 = x_1 + x_2$, right?

By Vieta's formula we have that
$\displaystyle x_1 x_2 = \dfrac{c}{a}$
and
$\displaystyle x_1 + x_2 = - \dfrac{b}{a}$

Thus $\displaystyle \dfrac{c}{a} = -\dfrac{b}{a}$.

Clearly then, c = -b, so your quadratic will be $\displaystyle y = a x^2 + bx - b$. Your example is of the same form.

-Dan
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September 28th, 2019, 04:05 PM   #5
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Quote:
Originally Posted by idontknow View Post
Find the quadratic equation which satisfies the condition $\displaystyle x_1 x_2 =x_1 +x_2$.
The equation $x_1x_2 = x_1 + x_2$ Is equivalent to $x_1x_2 - x_1 - x_2 = 0$ and the left side of this is already a quadratic polynomial so I have no clue what you are asking.
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