# Quadratic Equation Formula For Class 10th With Examples The quadratic equation is an important concept in mathematics, particularly in algebra. It is a second-degree polynomial equation in a single variable.

The quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and “x” represents the variable. The solutions to the quadratic equation are the values of “x” that satisfy the equation. The general formula to solve the quadratic equation is:

x = (-b ± √(b^2 – 4ac)) / (2a)

This formula is derived from the quadratic formula. Here are the individual components of the formula:

1. Discriminant: The discriminant is the expression under the square root in the quadratic formula. It is given by Δ = b^2 – 4ac. The value of the discriminant determines the nature of the solutions.
2. Square root: The symbol √ represents the square root of a number.
3. Coefficients: “a,” “b,” and “c” are the coefficients of the quadratic equation.
4. Plus-minus sign (±): This symbol indicates that there are two possible solutions, one with a positive sign and the other with a negative sign.

By substituting the values of “a,” “b,” and “c” into the quadratic formula and performing the necessary calculations, you can determine the solutions to the quadratic equation.

It’s important to note that the quadratic equation may have two real solutions, one real solution (when the discriminant is zero), or two complex solutions (when the discriminant is negative).

Using the quadratic formula, you can find the solutions to quadratic equations and solve problems involving quadratic functions and equations in mathematics and various fields of science and engineering.

The quadratic equation can have zero, one, or two real solutions, depending on the values of a, b, and c. To find these solutions, we can use the quadratic formula, which states that the solutions of the equation ax^2 + bx + c = 0 are given by:

x = (-b ± v(b^2 – 4ac)) / (2a)

Let’s break down the components of this formula and understand how it works.

The quadratic formula is derived from completing the square, which is a method to solve quadratic equations. It provides a systematic approach to finding the solutions, regardless of the values of a, b, and c.

To understand how the quadratic formula is derived, let’s start with a quadratic equation in the general form: ax^2 + bx + c = 0.

Step 1: Divide the equation by ‘a’ to make the coefficient of x^2 equal to 1:
x^2 + (b/a)x + c/a = 0

Step 2: Move the constant term (c/a) to the right-hand side:
x^2 + (b/a)x = -c/a

Step 3: Add the square of half the coefficient of x to both sides to complete the square. The coefficient of x is (b/2a), so we add (b/2a)^2 to both sides:
x^2 + (b/a)x + (b/2a)^2 = -c/a + (b/2a)^2

Simplifying the right-hand side gives:
x^2 + (b/a)x + (b^2/4a^2) = (b^2 – 4ac) / (4a^2)

Step 4: Rewrite the left-hand side as a perfect square:
(x + b/2a)^2 = (b^2 – 4ac) / (4a^2)

Step 5: Take the square root of both sides:
x + b/2a = ±v((b^2 – 4ac) / (4a^2))

Step 6: Solve for x by isolating it on one side:
x = (-b ± v(b^2 – 4ac)) / (2a)

This is the quadratic formula we mentioned earlier. Now, let’s use this formula to solve quadratic equations.

Example 1: Solve the equation 2x^2 – 5x + 2 = 0 using the quadratic formula.

In this equation, a = 2, b = -5, and c = 2. Plugging these values into the quadratic formula, we get:
x = (-(-5) ± v((-5)^2 – 4 * 2 * 2)) / (2 * 2)
= (5 ± v(25 – 16)) / 4
= (5 ± v9) / 4
= (5 ± 3) / 4

So, the solutions are:
x = (5 + 3) / 4 = 8 / 4 = 2
x = (5 – 3) / 4 = 2 / 4 = 0.5

Therefore, the solutions to the equation 2x^2 – 5x + 2 = 0 are x = 2 and x = 0.5.

Example 2: Solve the equation x^2 + 4x + 4 = 0 using the quadratic formula.

In this equation, a = 1, b = 4, and c = 4. Substituting these values into the quadratic formula, we have:
x = (-4 ± v(4^2 – 4 * 1 * 4)) / (2 * 1)
= (-4 ± v(16 – 16)) / 2
= (-4 ± v0) / 2
= (-4 ± 0) / 2

Since the discriminant (the term inside the square root) is zero, the solutions are identical. Hence, we have:
x = (-4 + 0) / 2 = -4 / 2 = -2

Therefore, the solution to the equation x^2 + 4x + 4 = 0 is x = -2.

Example 3: Solve the quadratic equation x^2 – 5x + 6 = 0

Factoring gives (x – 2)(x – 3) = 0

Setting each factor equal to zero: x – 2 = 0 or x – 3 = 0

Solving for “x”: x = 2 or x = 3

The solutions to the equation are x = 2 and x = 3.

Example 4: Solve the quadratic equation 2x^2 + 3x – 2 = 0

Applying the quadratic formula: x = (-3 ± √(3^2 – 42(-2))) / (2*2)

Simplifying: x = (-3 ± √(9 + 16)) / 4 x = (-3 ± √25) / 4 x = (-3 ± 5) / 4

Solving for “x”: x = (-3 + 5) / 4 or x = (-3 – 5) / 4 x = 2/4 = 1/2 or x = -8/4 = -2

The solutions to the equation are x = 1/2 and x = -2

In summary, the quadratic equation is an essential concept in algebra. It allows us to find the solutions to quadratic equations using the quadratic formula. By plugging in the values of a, b, and c into the formula, we can solve for x and determine the roots of the quadratic equation. Understanding and applying the quadratic formula is crucial for solving various mathematical problems and real-life applications.