Sreedhar acharya s formulators

          Sridharacharya Method is used to find solutions to quadratic equations of the form ax2 + bx + c = 0, a ≠ 0 and is given by x = (-b ± √(b2 - 4ac)) / 2a..

          By using the Sridhar Acharya formula, the solution for the above quadratic equation reduces to: urn:x-wileymedia:jgra

        1. By using the Sridhar Acharya formula, the solution for the above quadratic equation reduces to: urn:x-wileymedia:jgra
        2. Dr.
        3. Sridharacharya Method is used to find solutions to quadratic equations of the form ax2 + bx + c = 0, a ≠ 0 and is given by x = (-b ± √(b2 - 4ac)) / 2a.
        4. Psychosis, a chronic mental illness, can be treated with plants and traditional herbs, which contain phytochemicals and counter oxidants to.
        5. The current study aims to develop UFLC methods for 6-gingerol, biflorin, vasicinone, pellitorine, costunolide, dehydrocostuslactone, piperine, and apigenin in.

          6. Methods of Solving Quadratic Equations

          We will discuss here about the methods of solving quadratic equations.

          The quadratic equations of the form ax\(^{2}\) + bx + c = 0 is solved by any one of the following two methods (a) by factorization and (b) by formula.

          (a) By factorization method:

          In order to solve the quadratic equation ax\(^{2}\) + bx + c = 0, follow these steps:

          Step I: Factorize ax\(^{2}\) + bx + c in linear factors by breaking the middle term or by completing square.

          Step II: Equate each factor to zero to get two linear equations (using zero-product rule).

          Step III: Solve the two linear equations.

          This gives two roots (solutions) of the quadratic equation.

          Quadratic equation in general form is

          ax\(^{2}\) + bx + c = 0, (where a ≠  0) ………………… (i)

          Multiplying both sides of, ( i) by 4a,

          4a\(^{2}\)x\(^{2}\) + 4abx + 4ac = 0

          ⟹ (2ax)\(^{2}\) + 2 .

          2ax . b + b\(^{2}\) + 4ac - b\(^{2}\) = 0

          ⟹ (2ax + b)\(^{2}\) = b\(^{2}\) - 4ac [on sim