Quadratic Equations Concept, Formula and Free Online Practice Tests

Quadratic Equations

Quadratic Equations are polynomial equations of degree 2 in one variable, such as $$f(x) = ax^2 + bx + c = 0, $$where a, b, and c are all in R and a is not zero. It is the general form of a quadratic equation, with ‘a’ denoting the leading coefficient and ‘c’ denoting the constant term of f (x). The quadratic equation’s roots are the values of x that satisfy the equation (Ξ±, Ξ²).

Quadratic Equation Formula

The quadratic formula determines the solution or roots of a quadratic problem.

$$ x = (Ξ±, Ξ²) = \frac{-bΒ±\sqrt {b^2-4ac }}{2a} $$Β 

How Do You Solve Quadratic Equations?

There are generally four ways to solve quadratic problems. They are:

  1. Factoring
  2. Complete the square.
  3. Using the Quadratic Formula
  4. Taking the Square Root
Quadratic Equations Quizzes

Quadratic Equations Quiz 1

Quadratic Equations Quiz 2

Quadratic Equations Quiz 3

Quadratic Equations Questions

Q1. (i) 2xΒ² + 9x + 9 = 0
(ii) 15yΒ² + 16y + 4 = 0

(a) x > y
(b) x < y
(c) x β‰₯ y
(d) x ≀ y
(e) x = y or no relation can be established between x & y.

Ans.(b)
Sol.
(i) 2xΒ² + 9x + 9 = 0
2xΒ² + (6 + 3) x + 9 = 0
2x (x + 3) + 3 (x + 3) = 0
x = –3/2 , –3
(ii) 15yΒ² + 16y + 4 = 0
15yΒ² + 10y + 6y + 4 = 0
5y (3y + 2) + 2 (3y + 2) = 0
y = –2/5, –2/3
x < y

Q2. (i) 2x³ = √256
(ii) 2yΒ² – 9y + 10 = 0

(a) x = y or no relation can be established between x & y.
(b) x < y
(c) x ≀ y
(d) x β‰₯ y
(e) x > y

Ans.(c)
Sol.
(i) 2xΒ³ = 16
xΒ³ = 8
x = 2
(ii) 2yΒ² – 9y + 10 = 0
2yΒ² – (5 + 4) y + 10 = 0
2yΒ² – 5y – 4y + 10 = 0
y (2y – 5) – 2 (2y – 5) = 0
y = 2, 5/2
x ≀ y

Q3. (i) 6xΒ² – 11x + 4 = 0
(ii) 3yΒ² – 5y + 2 = 0

(a) x ≀ y
(b) x < y
(c) x β‰₯ y
(d) x > y
(e) x = y or no relation can be established between x & y.

Ans.(e)
Sol.
(i) 6xΒ² – 11x + 4 = 0
6xΒ² – (8 + 3) x + 4 = 0
6xΒ² – 8x – 3x + 4 = 0
2x (3x – 4) – 1 (3x – 4) = 0
x = 12, 43
(ii) 3yΒ² – 5y + 2 = 0
3yΒ² – (3 + 2) y + 2 = 0
3yΒ² – 3y – 2y + 2 = 0
3y (y – 1) – 2 (y – 1) = 0
y = β…”, 1
No relation between x and y

Q4. (i) 3xΒ² + 11x + 10 = 0
(ii) 2yΒ² + 11y + 14 = 0

(a) x β‰₯ y
(b) x ≀ y
(c) x > y
(d) x < y
(e) x = y or no relation can be established between x & y.

Ans.(a)
Sol.
(i) 3xΒ² + 11x + 10 = 0
3xΒ² + 6x + 5x + 10 = 0
3x (x + 2) + 5 (x + 2) = 0
x = – 2, –5/3
(ii) 2yΒ² + 11y + 14 = 0
2yΒ² + 7y + 4y + 14 = 0
y (2y + 7) +2 (2y + 7) = 0
y = –2, –7/2
x β‰₯ y

Q5. (i) 12xΒ² + 11x + 2 = 0
(ii) 12yΒ²+ 7y + 1 = 0

(a) x β‰₯ y
(b) x = y or no relation can be established between x & y.
(c) x < y
(d) x ≀ y
(e) x > y

Ans.(b)
Sol.
(i) 12xΒ² + 8x + 3x + 2 = 0
4x (3x + 2) + 1 (3x + 2) = 0
x = –2/3,–1/4
(ii) 12yΒ² + 7y + 1 = 0
12yΒ² + 4y + 3y + 1 = 0
4y (3y + 1) +1 (3y + 1) = 0
y = –1/3,–1/4
No relation between x and y

Q6. (i) 21xΒ² + 10x + 1 = 0
(ii) 24yΒ²+ 26y + 5 = 0

(a) x ≀ y
(b) x = y or no relation can be established between x & y.
(c) x β‰₯ y
(d) x > y
(e) x < y

Ans.(b)
Sol.
(i) 21xΒ² + 10x + 1 = 0
21xΒ² + 7x + 3x + 1 = 0
7x (3x + 1) + 1 (3x + 1) = 0
x = –1/3, –1/7
(ii) 24yΒ² + 26y + 5 = 0
24yΒ² + (20 + 6)y + 5 = 0
24yΒ² + 20y + 6y + 5 = 0
4y (6y + 5) + 1 (6y + 5) = 0
y = –5/6, –1/4
No relation between x and y.

Q7.
I. xΒ² – 7x + 12 = 0
II. yΒ² – 8y + 12 = 0

(a) If x > y
(b) If x β‰₯ y
(c) If y > x
(d) If y β‰₯ x
(e) If x = y or no relation can be established

Ans.(e)
Sol.
I. xΒ² – 7x + 12 = 0
xΒ²βˆ’4π‘₯βˆ’3π‘₯+12=0
(π‘₯βˆ’4)(π‘₯βˆ’3)=0 π‘₯=3,4
II. yΒ² – 8y + 12 = 0
yΒ²βˆ’6π‘¦βˆ’2𝑦+12=0
(π‘¦βˆ’6)(π‘¦βˆ’2)=0 𝑦=2,6
No relation can be established

Q8.
I. 2xΒ² + x – 28 = 0
II. 2yΒ² – 23y + 56 = 0

(a) If x > y
(b) If x β‰₯ y
(c) If y > x
(d) If y β‰₯ x
(e) If x = y or no relation can be established

Ans.(d)
Sol.
I. 2xΒ² + x – 28 = 0
2xΒ² + 8x – 7x – 28 = 0
2x (x + 4) – 7 (x + 4) = 0
(2x – 7) (x + 4)= 0
π‘₯=βˆ’4,7/2

II. 2yΒ² – 23y + 56 = 0
2yΒ² – 16y – 7y + 56 = 0
2y(y – 8) – 7(y – 8) = 0
(2y – 7) (y – 8) = 0
𝑦=7/2,8
y β‰₯ x

Q9.
I. 2xΒ² – 7x – 60 = 0
II. 3yΒ² + 13y + 4 = 0

(a) If x > y
(b) If x β‰₯ y
(c) If y > x
(d) If y β‰₯ x
(e) If x = y or no relation can be established

Ans.(e)
Sol.
I. 2xΒ² – 7x – 60 = 0
2xΒ² – 15x + 8x – 60 = 0
x (2x – 15 ) + 4 (2x – 15) = 0
(x + 4) (2π‘₯βˆ’15)=0
π‘₯= βˆ’4,15/2

II. 3yΒ² + 13y + 4 = 0
3yΒ² + 12y + y + 4 = 0
3y (y + 4) + 1 (y + 4) = 0
(3y + 1) (y + 4) = 0
𝑦=βˆ’13,βˆ’4
No relation between x and y

Q10.
I. xΒ² – 17x – 84 = 0
II. yΒ² + 4y – 117 = 0

(a) If x > y
(b) If x β‰₯ y
(c) If y > x
(d) If y β‰₯ x
(e) If x = y or no relation can be established

Ans.(e)
Sol.
I. xΒ² – 17x – 84 = 0
xΒ² +4x – 21x – 84 = 0
(x + 4) (x – 21) = 0
x = -4, 21
II. yΒ² + 4y – 117 = 0
yΒ² – 9y + 13y – 117 = 0
(y – 9) (y + 13) = 0
y = 9, -13
No relation between x and y


Latest Job Notifications | Daily Job updates

Frequently Asked Questions

Q1.What is a quadratic equation?
A quadratic equation is a polynomial equation with the maximum degree equal to two. The equation is provided by axΒ² + bx + c = 0, where an is not zero.

Q2: What are the procedures for solving a quadratic equation?
There are four main approaches for solving quadratic equations. They are Factorisation, Using Square Roots, Complete the square and Using Quadratic Formula.

Q3.Write the quadratic equation as a sum and product of roots.
If Ξ± and Ξ² are the roots of a quadratic equation, their total is Ξ±+Ξ².
The product of roots equals Ξ±Ξ².
Thus, the required equation is: x^2 – (Ξ±+Ξ²)x + (Ξ±Ξ²) = 0.