Using the roots of the equation find the value of a & b

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Discussion Overview

The discussion revolves around finding the values of variables \(a\) and \(b\) from the roots of the equation \((x+a)(x-b)=0\), given that the roots are -3 and 2. The scope includes mathematical reasoning and exploration of different approaches to solving the problem.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • Some participants suggest that the roots of the equation are -a and b, leading to the equations -a = -3 and b = 2, or b = -3 and -a = 2.
  • Another participant elaborates on the quadratic form of the equation, using the quadratic formula to derive the roots and arrives at the same pairs for \(a\) and \(b\): either \(a = 3\) and \(b = 2\) or \(a = -2\) and \(b = -3\).

Areas of Agreement / Disagreement

Participants present multiple approaches to derive the values of \(a\) and \(b\), but there is no consensus on a single method or final answer, as different interpretations of the roots lead to different pairs of values.

Contextual Notes

The discussion does not resolve the assumptions regarding the unordered nature of the roots and how they relate to the values of \(a\) and \(b\). There are also unresolved steps in the mathematical reasoning presented.

mathlearn
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Hello everybody after a little while :D

The roots of the equation $(x+a) (x-b)= 0$ are -3 or 2. Find the value of $a$ & $b$

What should I do here ?

Many Thanks :)
 
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mathlearn said:
Hello everybody after a little while :D

The roots of the equation $(x+a) (x-b)= 0$ are -3 or 2. Find the value of $a$ & $b$

What should I do here ?

Many Thanks :)

what are the roots of equation $(x+a) (x-b)= 0$
they are -a and b. this is unordered pair

so -a = -3 , b = 2 or b = -3 and -a = 2
 
kaliprasad said:
what are the roots of equation $(x+a) (x-b)= 0$
they are -a and b. this is unordered pair

so -a = -3 , b = 2 or b = -3 and -a = 2

Thanks :D
 
The hard way: (x+ a)(x- b)= x^2+ (a- b)x- ab. By the quadratic formula, the roots of that equation are \frac{b- a\pm\sqrt{(a- b)^2- (-4)ab}}{2}= \frac{b- a\pm\sqrt{a^2- 2ab+ b^2+ 4ab}}{2a}= \frac{b- a\pm\sqrt{a^2+ 2ab+ b^2}}{2}= \frac{b- a\pm\sqrt{(a+ b)^2}}{2}= \frac{b- a\pm (a+ b)}{2} so x= \frac{b- a+ a+ b}{2}= b or x= \frac{b- a- a+ b}{2}= -a

Since we are told that the two roots are -3 and 2 we can have either a= -(-3)= 3 and b= 2 or a= -2 and b= -3.
 

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