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:
- Factoring
- Complete the square.
- Using the Quadratic Formula
- Taking the Square Root
| Quadratic Equations Quizzes | ||
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 β€ 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 = β3/2 , β3
(ii) 15yΒ² + 16y + 4 = 0
15yΒ² + 10y + 6y + 4 = 0
5y (3y + 2) + 2 (3y + 2) = 0
y = β2/5, β2/3
x < y
Q2. (i) 2xΒ³ = β256
(ii) 2yΒ² β 9y + 10 = 0
(a) x = y or no relation can be established between x & y.
(b) x < y
(c) x β€ y
(d) x β₯ y
(e) x > y
Ans.(c)
Sol.
(i) 2xΒ³ = 16
xΒ³ = 8
x = 2
(ii) 2yΒ² β 9y + 10 = 0
2yΒ² β (5 + 4) y + 10 = 0
2yΒ² β 5y β 4y + 10 = 0
y (2y β 5) β 2 (2y β 5) = 0
y = 2, 5/2
x β€ y
Q3. (i) 6xΒ² β 11x + 4 = 0
(ii) 3yΒ² β 5y + 2 = 0
(a) x β€ 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 β 11x + 4 = 0
6xΒ² β (8 + 3) x + 4 = 0
6xΒ² – 8x β 3x + 4 = 0
2x (3x β 4) β 1 (3x β 4) = 0
x = 12, 43
(ii) 3yΒ² β 5y + 2 = 0
3yΒ² β (3 + 2) y + 2 = 0
3yΒ² β 3y β 2y + 2 = 0
3y (y β 1) β 2 (y β 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 β€ 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 = β 2, β5/3
(ii) 2yΒ² + 11y + 14 = 0
2yΒ² + 7y + 4y + 14 = 0
y (2y + 7) +2 (2y + 7) = 0
y = β2, β7/2
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 β€ y
(e) x > y
Ans.(b)
Sol.
(i) 12xΒ² + 8x + 3x + 2 = 0
4x (3x + 2) + 1 (3x + 2) = 0
x = β2/3,β1/4
(ii) 12yΒ² + 7y + 1 = 0
12yΒ² + 4y + 3y + 1 = 0
4y (3y + 1) +1 (3y + 1) = 0
y = β1/3,β1/4
No relation between x and y
Q6. (i) 21xΒ² + 10x + 1 = 0
(ii) 24yΒ²+ 26y + 5 = 0
(a) x β€ 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 = β1/3, β1/7
(ii) 24yΒ² + 26y + 5 = 0
24yΒ² + (20 + 6)y + 5 = 0
24yΒ² + 20y + 6y + 5 = 0
4y (6y + 5) + 1 (6y + 5) = 0
y = β5/6, β1/4
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
xΒ²β4π₯β3π₯+12=0
(π₯β4)(π₯β3)=0 π₯=3,4
II. yΒ² – 8y + 12 = 0
yΒ²β6π¦β2π¦+12=0
(π¦β6)(π¦β2)=0 π¦=2,6
No relation can be established
Q8.
I. 2xΒ² + x β 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 β 28 = 0
2xΒ² + 8x β 7x β 28 = 0
2x (x + 4) β 7 (x + 4) = 0
(2x β 7) (x + 4)= 0
π₯=β4,7/2
II. 2yΒ² – 23y + 56 = 0
2yΒ² – 16y β 7y + 56 = 0
2y(y β 8) β 7(y β 8) = 0
(2y β 7) (y β 8) = 0
π¦=7/2,8
y β₯ x
Q9.
I. 2xΒ² – 7x β 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 β 60 = 0
2xΒ² – 15x + 8x β 60 = 0
x (2x β 15 ) + 4 (2x β 15) = 0
(x + 4) (2π₯β15)=0
π₯= β4,15/2
II. 3yΒ² + 13y + 4 = 0
3yΒ² + 12y + y + 4 = 0
3y (y + 4) + 1 (y + 4) = 0
(3y + 1) (y + 4) = 0
π¦=β13,β4
No relation between x and y
Q10.
I. xΒ² – 17x β 84 = 0
II. yΒ² + 4y β 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 β 84 = 0
xΒ² +4x β 21x β 84 = 0
(x + 4) (x β 21) = 0
x = -4, 21
II. yΒ² + 4y β 117 = 0
yΒ² – 9y + 13y β 117 = 0
(y β 9) (y + 13) = 0
y = 9, -13
No relation between x and y
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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.