# Case 1

If Î± and Î² are the roots of the equation

Or,
In general, if

Then

*ax*^{2}+*bx*+*c*= 0, then we can write*ax*^{2}+*bx*+*c*=*x*^{2}â€“ (Î± + Î²)*x*+ Î± Î² =*x*^{2}â€“ (sum of the roots)*x*+ product of the roots = 0Or,

*ax*^{2}+*bx*+*c*=*a*(*x*âˆ’ Î±)(*x*âˆ’ Î²) = 0*x*_{1},*x*_{2},*x*_{3},â€¦,*x*_{n}_{ }are the roots of the equation*f*(*x*) =*ax*+^{n}*bx*^{n}^{-}^{1}+*cx*^{n}^{-}^{2}+â€¦+ K = 0Then

*f*(*x*) =*ax*+^{n}*bx*^{n}^{-}^{1}+*cx*^{n}^{-}^{2}+â€¦+ K =*a*(*x*âˆ’*x*_{1}) (*x*â€“*x*_{2}) (*x*â€“*x*_{3})â€¦(*x*â€“*x*)_{n}

# Case 2

*p*and

*q*are the roots of the equation

*ax*

^{2}+

*bx*+

*c*= 0, and the quadratic equation with (1/

*p*) and (1/

*q*) as the roots is to be found:

Put
or,
So, the quadratic equation is:

*x*= 1/*p**a*(1/*p*)^{2}+*b*(1/*p*) +*c*= 0*cp*^{2}+*bp*+*a*= 0*cx*^{2}+*bx*+*a*= 0

# Case 3

*p*and

*q*are the roots of the equation

*ax*

^{2}+

*bx*+

*c*= 0, and the quadratic equation with (âˆ’

*p*) and (âˆ’

*q*) as the roots is to be found:

Put
Or,
So, the quadratic equation is:

*x*= âˆ’*p**a*(âˆ’*p*)^{2}+*b*(âˆ’*p*) +*c*= 0*cp*^{2}âˆ’*bp*+*a*= 0*cx*^{2}âˆ’*bx*+*a*= 0Example

What will be the quadratic equation having the roots opposite in sign as that of the quadratic equation

*x*^{2}+ 5*x*+ 6 = 0?Solution

Putting (â€“
Desired equation is (âˆ’
So, equation is:
Alternatively, the roots of the quadratic equation
So, the roots of the desired equation should be 2 and 3.
So, the quadratic equation is (

*x*) at the place of*x*will give us the solution.*x*)^{2}+ 5 (âˆ’*x*) + 6 = 0*x*^{2}âˆ’ 5*x*+ 6 = 0*x*^{2}+ 5*x*+ 6 = 0 are â€“2 and â€“3.*x*âˆ’2) (*x*âˆ’3) = 0, or,*x*^{2}âˆ’5*x*+ 6 = 0

# Case 4

*p*and

*q*are the roots of the equation

*ax*

^{2}+

*bx*+

*c*= 0, and the quadratic equation with (

*p*)

^{2}and (

*q*)

^{2}as the roots is to be found:

Let
Solving this will give us the desired equation.

*x*be a root of the original equation and*y*be a root of the transformed equation.*y*=*x*^{2}, so we will put*x*= âˆš*y**a*(âˆš*y*)^{2}+*b*(âˆš*y*) +*c*= 0,