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

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SUMMARY

The equation $(x+a)(x-b)=0$ has roots -3 and 2, leading to the determination of values for a and b. The roots can be expressed as -a and b, resulting in two possible pairs: a = 3, b = 2 or a = -2, b = -3. The quadratic formula confirms these roots through the transformation of the equation into standard form, validating the derived values for a and b. This analysis provides a clear method for solving similar equations involving roots.

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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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