What is the Sridharacharya Formula?

Quadratic equations are equations of the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are constants and $x$ is the variable. The Sridharacharya formula is a method for solving quadratic equations that was developed by the Indian mathematician Sridhara in the 12th century.

The Sridharacharya Formula

The Sridharacharya formula states that the solutions to the quadratic equation $ax^2 + bx + c = 0$ are given by:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where:

• $a$, $b$, and $c$ are the constants from the quadratic equation
• $x$ is the variable

How to Use the Sridharacharya Formula

To use the Sridharacharya formula, simply substitute the values of $a$, $b$, and $c$ into the formula and solve for $x$.

For example, to solve the quadratic equation $2x^2 + 3x - 5 = 0$, we would substitute $a = 2$, $b = 3$, and $c = -5$ into the formula:

$$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)}$$

Simplifying this expression, we get:

$$x = \frac{-3 \pm \sqrt{9 + 40}}{4}$$

$$x = \frac{-3 \pm \sqrt{49}}{4}$$

$$x = \frac{-3 \pm 7}{4}$$

So the solutions to the quadratic equation $2x^2 + 3x - 5 = 0$ are $x = 1$ and $x = -5/2$.

Advantages of the Sridharacharya Formula

The Sridharacharya formula is a simple and straightforward method for solving quadratic equations. It is also very efficient, and it can be used to solve quadratic equations with any coefficients.

Disadvantages of the Sridharacharya Formula

The Sridharacharya formula can be difficult to remember, and it can be easy to make mistakes when using it. Additionally, the formula does not work for quadratic equations that have complex solutions.

The Sridharacharya formula is a powerful tool for solving quadratic equations. It is simple to use and very efficient, but it can be difficult to remember and it does not work for quadratic equations with complex solutions.

Solved Examples Using Sridharacharya Formula

Example 1

Solve the quadratic equation $x^2 - 4x - 5 = 0$ using the Sridharacharya formula.

Solution:

Comparing the given equation with the standard quadratic equation $ax^2 + bx + c = 0$, we have $a = 1$, $b = -4$, and $c = -5$. Substituting these values into the Sridharacharya formula, we get:

$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}$$

Simplifying the expression, we get:

$$x = \frac{4 \pm \sqrt{16 + 20}}{2}$$

$$x = \frac{4 \pm \sqrt{36}}{2}$$

$$x = \frac{4 \pm 6}{2}$$

Therefore, the solutions to the equation $x^2 - 4x - 5 = 0$ are $x = 5$ and $x = -1$.

Example 2

Solve the quadratic equation $2x^2 + 3x - 5 = 0$ using the Sridharacharya formula.

Solution:

Comparing the given equation with the standard quadratic equation $ax^2 + bx + c = 0$, we have $a = 2$, $b = 3$, and $c = -5$. Substituting these values into the Sridharacharya formula, we get:

$$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)}$$

Simplifying the expression, we get:

$$x = \frac{-3 \pm \sqrt{9 + 40}}{4}$$

$$x = \frac{-3 \pm \sqrt{49}}{4}$$

$$x = \frac{-3 \pm 7}{4}$$

Therefore, the solutions to the equation $2x^2 + 3x - 5 = 0$ are $x = 1$ and $x = -\frac{5}{2}$.