CBSE Class 10 Mathematics Chapter 4 - Quadratic Equations - Quadratic Formula

 CBSE Class 10 Mathematics  Chapter 4

 - Quadratic Equations - Quadratic Formula

Quadratic Formula

• To find the solutions (roots) of a quadratic equation in the form ax² + bx + c = 0, we can use the quadratic formula:

The ± symbol indicates that there are generally two solutions: one for the plus sign and one for the minus sign.

Discriminant (D)

• The value inside the square root in the quadratic formula, b² - 4ac, is called the discriminant. It determines the nature of the roots:
• If the discriminant is positive, there are two real and distinct roots.
• If it is zero, there is one real and repeated root (a perfect square).
• If it is negative, there are two complex (non-real) roots.

Completing the Square Method

Step 1: Ensure the Equation is in Standard Form

Make sure the quadratic equation is in the standard form: ax² + bx + c = 0, where a≠0.

Step 2: Move the Constant Term to the Other Side

If the constant term 'c' is on the left side of the equation, move it to the right side by subtracting 'c' from both sides. This ensures that the right side of the equation is equal to zero:

ax² + bx = -c

Step 3: Divide by 'a' (if 'a' is not 1)

If 'a' is not equal to 1, divide both sides of the equation by 'a' to make the coefficient of the x² term equal to 1. This simplifies the equation:

x² + (b/a)x = -c/a

Step 4: Create a Perfect Square Trinomial

To complete the square, focus on the terms involving 'x.' We want to create a perfect square trinomial on the left side. To do this, take half of the coefficient of 'x' (which is (b/a)) and square it. Add and subtract this value on the left side:

x² + (b/a)x + (b/2a)² - (b/2a)² = -c/a

Step 5: Simplify the Left Side

On the left side, we now have a perfect square trinomial:

(x + b/2a)² - (b/2a)² = -c/a

Step 6: Solve for x

Isolate the perfect square trinomial on the left side and simplify the right side:

(x + b/2a)² = (b² - 4ac) / (4a²)

Step 7: Take the Square Root

Take the square root of both sides to solve for 'x':

x + b/2a = ±√(b² - 4ac) / (2a)

Step 8: Solve for 'x'

Isolate 'x' on the left side by subtracting b/2a from both sides:

x = (-b ± √(b² - 4ac)) / (2a)

Step 9: Finalize the Solutions

We have the solutions for 'x'. If the discriminant (b² - 4ac) is positive, we will have two distinct real solutions (±). If it's zero, we will have one real solution and if it's negative, we will have two complex solutions.