Maths Guys

QUADRATIC FORMULA (SHRIDHARACHARYA SUTRA) – The quadratic formula is given by the great Indian mathematician SHRIDHARACHARYA and is also known as SHRIDHARACHARYA SUTRA. This formula is used to solve the quadratic equation.

x = –b ± √(b^2 – 4ac)/2a

Where, x = variable

a, b, c = Coefficients of quadratic equation

This formula can be derived from the perfect square method.

The standard form of a quadratic equation is

ax^2 + bx + c = 0

x^2 + bx/a + c/a = 0 (taking 1 as coefficient of x2)

x^2 + bx/a = −c/a (taking constant term to right side)

x^2 + bx/a + (b/2a)2 = −c/a + (b/2a)2 (adding square of half of coefficient of x)

(x + b/2a)2 = −c/a + b^2/4a^2

(x + b/2a)2 = (−c⨯4a + b^2⨯1)/4a2 = (−4ac + b^2)/4a2

Taking square root both side,

√(x + b/2a)2 = √(−4ac + b2)/4a^2

x + b/2a = ±√(b^2 − 4ac)/2a

x = −b/2a ± √(b^2 − 4ac)/2a

x = −b ± √(b^2 − 4ac)/2a

Here sign ± shows the two values of x.

x = −b + √(b^2 − 4ac)/2a x = −b − √(b^2 − 4ac)/2a

The above values of x are the Quadratic Formula (Shridharacharya sutra).

Example – Solve the quadratic equation 9x*2 + 7x – 2 = 0 by Shridhar Acharya sutra.

Solution – given equation 9x^2 + 7x – 2 = 0

Comparing with standard form ax^2 + bx + c = 0

So, a = 9, b = 7, c = – 2

By quadratic formula, x = −b ± √(b^2 − 4ac)/2a

Putting the values of a, b, c

x = −7 ± √{72 − 4⨯9⨯(-2)}/2⨯9

x = −7 ± √{49 + 72}/18

x = −7 ± √{121}/18 = −7 ± 11/18

On taking the (+) sign,

x = −7 + 11/18

x = 4/18

x = 2/9

On taking the (−) sign,

x = −7 − 11/18

x = −18/18

x = −1

These values of x are the roots of the given quadratic equation. Ans.

DISCRIMINANT AND NATURE OF ROOTS
We can find the nature of roots by the term b2 – 4ac used in the Shridharacharya sutra which is called the Discriminant. it is denoted by D.

D = b^2 – 4ac

⇒ If D > 0 then the roots will be different and real.

x = (−b + √D)/2a and x = (−b − √D)/2a

⇒ If D = 0 then the roots will be equal and real.

x = −b/2a and x = −b/2a

⇒ If D < 0 then the roots will not be real, they will be imaginary.

SOME EXAMPLES –
Example – 1) find the nature of the roots of the equation 2x^2 – 6x + 3 = 0 and if roots exist then find them.

Solution – Given equation 2x^2 – 6x + 3 = 0

Comparing the equation with ax^2 + bx + c = 0

Here, a = 2, b = – 6, c = 3

Now D = b^2 – 4ac

D = (-6)2 – 4⨯2⨯3 = 36 – 24 = 12

∵ D > 0

∴ The roots of the equation 2x^2 – 6x + 3 = 0 will be different and Real.

By Shridharacharya formula, x = (−b ± √D)/2a where, D = b^2 – 4ac

x = {−(-6) ± √12)}/2⨯2 = {6 ± √(4⨯3)}/4

x = {6 ± 2√3}/4 = 2{3 ± √3}/4 = {3 ± √3}/2

So, the roots are x = (3 + √3)/2 and x = (3 − √3)/2. Ans.

Example – 2) find the nature of roots of the equation x^2 – 4x + 4 = 0 and if roots exist then find them.

Solution – Given equation x^2 – 4x + 4 = 0

By comparing with ax^2 + bx + c = 0

a = 1, b = – 4, c = 4

Now D = b^2 – 4ac

D = (-4)2 – 4⨯1⨯4 = 16 – 16 = 0

∵ D = 0

∴ The roots of the equation x^2 – 4x + 4 = 0 will be equal and Real.

By quadratic formula, x = (−b ± √D)/2a where, D = b^2 – 4ac

x = {−(-4) ± √0)/2⨯1 = (4 ± 0)/2 = 4/2

x = 2

So, the roots are x = 2 and x = 2. Ans.

Example – 3) find the nature of the roots of equation 2x2 + 3x + 5 = 0 and if roots exists then find them.

Solution – Given equation 2x2 + 3x + 5 = 0

On comparing with ax^2 + bx + c = 0

a = 2, b = 3, c = 5

Now D = b^2 – 4ac

D = (3)2 – 4⨯2⨯5 = 9 – 40 = –31

∵ D < 0

∴ The roots of the equation 2x2 + 3x + 5 = 0 will not be Real. They will be imaginary.

By Shridharacharya quadratic formula, x = (−b ± √D)/2a where, D = b2 – 4ac

x = (−3 ± √-31)/2⨯2 = (−3 ± √−1⨯31)/4

x = (−3 ± √−1⨯√31)/4 = (−3 ± √31i )/4

So, the roots x = (−3 + √31i )/4 and x = (−3 − √31i )/4. where, i(iota) = √−1