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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 September 28th, 2019, 06:32 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Quadratic polynomial Find the quadratic equation which satisfies the condition $\displaystyle x_1 x_2 =x_1 +x_2$. September 28th, 2019, 07:34 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 310 Thanks: 162 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? Thanks from topsquark and idontknow September 28th, 2019, 08:00 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 The equation is : $\displaystyle x^2 -2x+2=0$. It satisfies the condition $\displaystyle x_1 x_2 = x_1 + x_2$. September 28th, 2019, 08:28 AM   #4
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Quote:
 Originally Posted by idontknow 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 September 28th, 2019, 04:05 PM   #5
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Quote:
 Originally Posted by idontknow 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. Tags polynomial, quadratic Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ismael Calculus 2 August 23rd, 2014 05:27 PM mathLover Abstract Algebra 1 April 17th, 2012 02:19 AM panky Algebra 1 November 5th, 2011 07:09 AM condemath Algebra 3 September 20th, 2011 08:34 PM FelisCanisOfCadog Calculus 1 February 22nd, 2009 01:22 PM

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