Question

WHAT IS THE SHRIDHAR ACHARYA FORMULA

Answer 1

x=-b+-root b^2-4ac/2a


hope it helps you

Answer 2

Quadratic equation in general form is

ax22 + bx + c = 0, (where a ≠ 0) ………………… (i)

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

4a22x22 + 4abx + 4ac = 0

⟹ (2ax)22 + 2 . 2ax . b + b22 + 4ac - b22 = 0

⟹ (2ax + b)22 = b22 - 4ac [on simplification and transposition]

Now taking square roots on both sides we get

2ax + b = ±b2−4ac−−−−−−−√±b2−4ac)

⟹ 2ax = -b ±b2−4ac−−−−−−−√±b2−4ac)

⟹ x = −b±b2−4ac√2a−b±b2−4ac2a

i.e., −b+b2−4ac√2a−b+b2−4ac2a or, −b−b2−4ac√2a−b−b2−4ac2a

Solving the quadratic equation (i), we have got two values of x.

That means, two roots are obtained for the equation, one is x = −b+b2−4ac√2a−b+b2−4ac2a and the other is x = −b−b2−4ac√2a−b−b2−4ac2a



Example to Solving quadratic equation applying factorization method:

Solve the quadratic equation 3x22 - x - 2 = 0 by factorization method.

Solution:

3x22 - x - 2 = 0

Breaking the middle term we get,

⟹ 3x22 - 3x + 2x - 2 = 0

⟹ 3x(x - 1) + 2(x - 1) = 0

⟹ (x - 1)(3x + 2) = 0

Now, using zero-product rule we get,

x - 1 = 0 or, 3x + 2 = 0

⟹ x = 1 or x = -2323

Therefore, we get x = -2323, 1.

These are the two solutions of the equation