Quadratic Equations Solved Examples (Class 10 Maths)
Quadratic equations are polynomial equations of degree 2. They can be solved using factorization, completing the square, or the quadratic formula. Understa
TL;DR: Quadratic equations are polynomial equations of degree 2. They can be solved using factorization, completing the square, or the quadratic formula. Und…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
Quadratic equations are polynomial equations of degree 2. They can be solved using factorization, completing the square, or the quadratic formula. Understa
Quadratic Equations — Solved Numerical Examples (Step by Step)
Example 1: Solve the quadratic equation x^2 + 5x + 6 = 0 by factorization.
Solution: Given equation: x^2 + 5x + 6 = 0
We need to find two numbers that multiply to 6 and add to 5.
Those numbers are 2 and 3.
Factorizing: (x + 2)(x + 3) = 0
Setting each factor to zero:
x + 2 = 0 or x + 3 = 0
x = -2 or x = -3
Example 2: Solve the equation 2x^2 - 8x + 6 = 0 using the quadratic formula.
Solution: Given equation: 2x^2 - 8x + 6 = 0
Here a = 2, b = -8, c = 6
Quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a)
First calculate the discriminant:
b^2 - 4ac = (-8)^2 - 4(2)(6) = 64 - 48 = 16
sqrt(16) = 4
x = (8 ± 4) / 4
x = 12/4 = 3 or x = 4/4 = 1
Example 3: Solve x^2 - 4x + 4 = 0.
Solution: Given equation: x^2 - 4x + 4 = 0
This is a perfect square trinomial:
x^2 - 4x + 4 = (x - 2)^2
Setting to zero:
(x - 2)^2 = 0
x - 2 = 0
x = 2
Example 4: Find the nature of roots of 3x^2 - 6x + 4 = 0.
Solution: Given equation: 3x^2 - 6x + 4 = 0
Here a = 3, b = -6, c = 4
Calculate discriminant:
D = b^2 - 4ac
D = (-6)^2 - 4(3)(4)
D = 36 - 48
D = -12
Since D < 0, the equation has no real roots.
The roots are complex (imaginary).
Example 5: A rectangle has length 2 cm more than its width. If the area is 24 cm^2, find the dimensions.
Solution: Let width = x cm
Then length = (x + 2) cm
Area = length × width
24 = (x + 2) × x
24 = x^2 + 2x
x^2 + 2x - 24 = 0
Factoring: We need numbers that multiply to -24 and add to 2.
Those are 6 and -4.
(x + 6)(x - 4) = 0
x = -6 or x = 4
Since width cannot be negative, x = 4 cm
Width = 4 cm, Length = 6 cm
Example 6: Solve x^2 - 5x + 6 = 0 by completing the square.
Solution: Given equation: x^2 - 5x + 6 = 0
Rearrange: x^2 - 5x = -6
Complete the square:
x^2 - 5x + (5/2)^2 = -6 + (5/2)^2
x^2 - 5x + 25/4 = -6 + 25/4
(x - 5/2)^2 = -24/4 + 25/4
(x - 5/2)^2 = 1/4
Taking square root:
x - 5/2 = ±1/2
x = 5/2 + 1/2 = 3 or x = 5/2 - 1/2 = 2
Example 7: Check if x = 2 is a solution of 2x^2 - 5x + 2 = 0.
Solution: Substitute x = 2 in 2x^2 - 5x + 2:
2(2)^2 - 5(2) + 2
= 2(4) - 10 + 2
= 8 - 10 + 2
= 0
Since the result is 0, x = 2 is a solution.
Tips
- Try factorization first for simple quadratic equations with integer roots. Look for two numbers that multiply to c and add to b.
- Use the discriminant (b^2 - 4ac) to determine the nature of roots before solving: if D > 0 there are two real roots, D = 0 means one repeated root, D < 0 means no real roots.
- The quadratic formula always works but can involve fractions. Complete the square when you need to derive something or when the equation involves perfect squares.
Frequently Asked Questions
What does discriminant tell us?
The discriminant D = b^2 - 4ac tells us the nature of roots: if D > 0 the equation has two distinct real roots, if D = 0 it has one repeated real root, and if D < 0 it has no real roots (complex roots instead).
Can a quadratic equation have more than two roots?
No. By definition, a quadratic equation (degree 2) has at most 2 roots. They may be real and distinct, real and equal (repeated), or complex conjugates.
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