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Q1. What are the different ways to solve a quadratic equation?
Ans. Sometimes, solving quadratic equations can be quite difficult . However, various methods exist, depending on the nature of the quadratic equation. Four primary approaches are commonly employed: factoring, utilizing square roots, completing the square, and employing the quadratic formula.
Q2. What are the steps involved in factoring a quadratic equation?
Ans. To solve a quadratic equation using the factoring method, follow these steps:
Begin by consolidating all terms on one side of the equation, leaving zero on the opposite side, then proceed to factorize. Next, set each factor equal to zero and solve each resulting equation individually. Verify the solutions by substituting them back into the original equation.
Q3. What is the standard form of the quadratic equation?
Ans. The form ax2+bx+c=0 is said to be the standard form of a quadratic equation.
Quadratic Equation, Definition, Formula and Examples
Quadratic Equation can be defined as a quadratic polynomial equation, which means that it that it comprises a minimum of one term that is squared.
Priyanka Dahima3 Jul, 2024
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Key Concepts for Quadratic Equation:
A quadratic polynomial, when equated to zero, becomes a quadratic equation. In other terms, a quadratic equation is a second-degree algebraic equation. The values of x satisfying the equation are called the roots of the quadratic equation.
General form:
a
x
2
+bx+c=0
where
x
represents an unknown, and a, b, and c are constants with
a not equal to
0
. The constants a, b, and c are usually real numbers, but they can be complex numbers as well.
For Example
x
2
+5x+6=0
,
6
x
2
-17x+12=0
Quadratic Equation Formula
The Quadratic Formula expresses the solutions in terms of
a
,
b
, and
c
.
Best Quadratic Equation Examples
In each question, two equations will be given. You have to solve the equations and choose the appropriate relation between “x” and “y” and choose the appropriate relation between “x” and “y” from the options and choose the correct option.
a) x<y
b) x>y
c) xy
d) xy
e) x= y or the relation can not be established
1) I. x2-9x+20=0 II. y2-7y+12=0
Solution:
I. x
2
-9x+20=0
x
2
-4x-5x+20=0
x(x-4)-5(x-4)=0
x=4,5
II. y
2
-7y+12=0
y
2
-4y-3y+12=0
y(y-4)-3(y-4)=0
y=4,3
Case 1: x = 4, x = 4 (x = y)
Case 2: x = 4, x = 3 (x > y)
Case 3: x = 5, x = 4 (x > y)
Case 4: x = 5, x = 3 (x > y)
So,
x
y
2) I. x2-5x-14=0 II. y2-16y+64=0
Solution:
x
2
-5x-14=0
x
2
-7x+2x-14=0
x(x-7)+2(x-7)=0
x=7,-2
y
2
-16y+64=0
y
2
-8y-8y+64=0
y(y-8)-8(y-8)=0
y=8,8
Case 1: x = 7, y = 8 (y > x)
Case 2: x = 7, y = 8 (y > x)
Case 3: x = -2, y = 8 (y > x)
Case 4: x = -2, y = 8 (y > x)
So,
x<y
3) I. x2 + 7x+12=0 II. y2+9y+20=0
Solution:
I . x
2
+ 7x+12=0
x
2
+4x+3x +12=0
x(x
+ 4)+3(x+4) = 0
(x
+4)
(x
+3)=0
x =
-3, -4
y
2
+9y+20=0
y
2
+ 5y+4y+20=0
y(y
+ 5)+4
y+5
=0
y = -5, -4
Case 1: x = -3, y = -5 (x > y)
Case 2: x = -4, y = -5 (x > y)
Case 3: x = -3, y = -4 (x > y)
Case 4: x =-4, y = -4 (x = y)
So,
x
y
Tricks to Solve Quadratic Equations Questions:
Trick 1: Quadratic equations
solution based on the sign of coefficients in the equation:
For example:
x
2
-14x+45=0
y
2
+7y+6=0
In this example, the signs visible in the x equation are ( -, +) and signs visible in the y equation are (+, +). So, signs of actual value of x and y in the final answer will be ( +, +) and ( -, -)
So, “
Relationships can’t be established”.
Trick 2:
Constant term negative in both equations = Relationship can’t be established)
For example:
2
x
2
-11x-15=0
21
y
2
+23y-6=0
In this example, the constant term is negative in both equations. So, the correct answer is “
Relationships can’t be established”.
Weightage of Quadratic Equations in Banking Exams:
The weightage of quadratic equations can vary from one exam to another and even within different sections of the same exam. It's challenging to provide an exact weightage for quadratic equations, they are usually part of a broader set of topics in the quantitative aptitude section, which also includes topics like arithmetic, geometry, number systems, data interpretation, and so on. As such, candidates preparing for banking exams should ensure they have a solid understanding of quadratic equations and practice solving various types of problems to maximize their chances of success.
Roughly weightage - (0-5) Questions
Solved Problems:
1)
I. x
2
- 14x+45= 0
y
2
-18y+72=0
a)
x<y
b)
x>y
c)
x
y
d)
x
y
e) x= y or the relation can not be established
Solution:
I . x
2
- 14x+45= 0
x
2
- 9x-5x+45= 0
x(x
-9)-5(x-9)=0
x = 9, 5
y
2
-18y+72=0
y
2
– 12y-6y+72=0
y(y
– 12) -6
y-12
=0
y = 12, 6
Case 1: x = 9, y = 12 (y > x)
Case 2: x = 9 , y = 6 (x > y)
Case 3: x = 5, y = 12 (y > x)
Case 4: x = 5, y = 6 (y > x)
no relation can be established between x and y.
2) I. x2 + 32x + 240 = 0
y
2
+ 3y – 108 = 0
A) x > y
B) x < y
C) x = y or the relationship cannot be established
D) x ≥ y
E) x ≤ y
Solution:
From I:
x
2
+ 32x + 240 = 0
x
2
+ 20x + 12x + 240 = 0
x(x + 20) + 12(x + 20) = 0
(x + 20)(x + 12) = 0
x = -20, -12
From II:
y
2
+ 3y – 108 = 0
y
2
+ 12y – 9y – 108 = 0
y(y + 12) – 9(y + 19) = 0
(y + 12)(y – 9) = 0
y = -12, 9
Case 1: x = -20, y = -12 ( y > x)
Case 2: x = -20, y = 9 (y > x)
Case 3: x = -12, y = -12 ( x = y )
Case 4: x = -20, y = 9 (y > x)
So, y ≥ x
3) I. 6x² + 19x + 15 = 0
24y² + 11y + 1 = 0
a) if x > y
b) if x ≥ y
c) if x < y
d) if x ≤ y
e) if x = y or the relationship cannot be established.
⇒ 24y² + 8y + 3y + 1= 0
⇒ (3y + 1) (8y + 1) = 0
⇒ y = –1/3, –1/8
Case 1: x = -3/2, y = -⅓ (y > x)
Case 2: x = -3/2, y = -⅛ (y > x)
Case 3: x = -5/3, y = -⅓ (y > x)
Case 4: x = -5/3, y = -⅛ (y > x)
So, y > x
4) I. 2x2- x - 1=0
2y
2
- 4y +2=0
a) if x > y
b) if x ≥ y
c) if x < y
d) if x ≤ y
e) if x = y or the relationship cannot be established.
Solution:
From equation I
2x
2
-x-1=0
2
x
2
-2x+x-1=0
2x(x-1) +1(x-1)=0
(2x+1) (x-1)=0
x=
-1
2
, 1
From equation II
2y
2
- 4y +2=0
2y
2
- 2y - 2y +2=0
2y(y-1)- 2(y-1)=0
(2y-2) (y-1)=0
y = 1, 1
Case 1: x = -½, y = 1 (y>x)
Case 2: x = -½, y = 1 (y>x)
Case 3: x = 1, y = 1 (y=x)
Case 4: x = 1, y = 1 (y=x)
So, x ≤ y
5) I . 4x2- 7x - 57 =0
II. 5y
2
- 6y - 63=0
a) if x > y
b) if x ≥ y
c) if x < y
d)if x = y or the relationship cannot be established.
e) if x ≤ y
Solution:
From equation I
4x
2
-7x - 57 =0
4x
2
-19x+12x-57 =0
x(4x-19) +3(4x-19) =0
(4x-19) (x+3) =0
x=
19
4
,-3
From equation II
5y
2
- 6y - 63=0
5y
2
-21y+15y-63 =0
y(5y-21) +3(5y-21) =0
(y+3)(5y-21)=0
y = -3,
21
5
Case 1: x =
19
4
, y = -3 (x > y)
Case 2: x =
19
4
, y =
21
5
(x > y)
Case 3: x = -3, y = -3 (x = y)
Case 4: x = -3, y =
21
5
(y > x)
So, x = y or relationship between x and y can't be established.
Previous Year Questions Of Quadratic Equations Asked In Banking Exams :
1) I.
5
x
2
+3x-14=0
II.
5
y
2
+3y-2=0
2) I.
x
2
+9x+20=0
II.
7
y
2
-19y+10=0
3) I.
x
2
-7x+10=0
II.
y
2
-5y+6=0
4) I.
x
2
+12x+35=0
II.
3
y
2
+19y+20=0
5) I.
x
2
-3x-88=0
II.
y
2
+8y-48=0
6) I.
4
x
2
-13x+10=0
II.
2
y
2
-15y+28=0
7) I.
2
x
2
+15x+27=0
II.
3
y
2
+25y-18=0
8) I.
6
x
2
-5x+1=0
II.
15
y
2
-8y+1=0
9) I.
7
x
2
-23x+18=0
II.
4
y
2
-9y+5=
10) I.
10
x
2
-11x-6=0
II.
y
2
+5y+6=0
The Quadratic Formula expresses the solutions in terms of
For example:
