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

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The equation $(x+a)(x-b)=0$ has roots -3 and 2, leading to the relationships -a = -3 and b = 2, or b = -3 and -a = 2. Solving these gives two possible pairs: a = 3 and b = 2, or a = -2 and b = -3. The discussion emphasizes that the roots can be derived from the equation's structure, confirming the unordered nature of the pairs. Ultimately, the values of a and b can be determined through basic algebraic manipulation.
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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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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