Hello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating sample questions in Quadratic Equations which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!
Follow the link To solve Quadratic Equations with the help of Number Line
- x2 – 18x + 72= 0, 5y2 – 18y + 9 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedA) If X>Y
Explanation:
x2 – 18x + 72= 0
(x-12)(x-6) = 0
Gives x = 6, 12
5y2 – 18y + 9 = 0
5y2 – 15y – 3y + 9 = 0
Gives y = 3/5, 3
Put on number line
3/5 3 6 12 - x2 = 4, 3y2 – 4y – 4 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedE) If X=Y or cannot be established
Explanation:
x2 = 4
Gives x = 2 , -2
3y2 – 4y – 4 = 0
3y2 – 6y + 2y – 4 = 0
Gives y = -2/3, 2
Put on number line
-2 -2/3 2
When y = 2, y ≥ x
When y = -2/3, y > x(-2) and y < x(2)
So no relation - 6x2 – 5x – 6 = 0, 2y2 – 13y + 20 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedB) If X < Y
Explanation:
6x2 – 5x – 6 = 0
6x2 – 9x + 4x – 6 = 0
Gives x = -2/3, 3/2
2y2 – 13y + 20 = 0
2y2 – 8y – 5y +20 = 0
Gives y = 4, 5/2
Put on number line
-2/3 3/2 5/2 4 - 2x2 – 5x = 0, 2y2 + 7y – 4 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedE) If X=Y or cannot be established
Explanation:
2x2 – 5x = 0
x(2x-5) = 0
Gives x = 0, 5/2
3y2 – 7y – 6 = 0
3y2 – 9y + 2y – 6 = 0
Gives y = -2/3, 3
Put on number line
-2/3 0 5/2 3 - 2x2 + 5x + 2= 0, 2y2 + 19y + 45 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedA) If X > Y
Explanation:
2x2 + 5x + 2= 0
2x2 + 4x + x + 2= 0
Gives x = -1/2, -2
2y2 + 19y + 45 = 0
2y2 + 10y + 9y + 45 = 0
Gives y= -10/2, -9/2
Put on number line
-10/2 -9/2 -2 -1/2 - x2 + x – 20 = 0, 2y2 + 13y + 15 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedE) If X = Y or relation cannot be established
Explanation:
x2 + x – 20 = 0
(x+5)(x-4) = 0
Gives x = -5, 4
2y2 + 13y + 15 = 0
2y2 + 10y + 3y + 15 = 0
Gives y = -5, -3/2
Put on number line
-5 -3/2 4 - 5x2 – 7x – 6 = 0, 5y2 + 23y + 12 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedC) If X ≥ Y
Explanation:
5x2 – 7x – 6 = 0
5x2 – 10x + 3x – 6 = 0
Gives x = -3/5, 2
5y2 + 23y + 12 = 0
5y2 + 20y + 3y + 12 = 0
Gives y = -4, -3/5
Put on number line
-4 -3/5 2 - 3x2 + 8x + 5 = 0, 2y2 + y – 1 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedD) If X ≤ Y
Explanation:
3x2 + 8x + 5 = 0
3x2 + 3x + 5x + 5 = 0
Gives x = -1, -5/3
2y2 + y – 1 = 0
2y2 + 2y – y -1 = 0
Gives y = -1, 1/2
put on number line
-5/3 -1 1/2 - 2x + 5y = 23.5, 5x+ 2y = 22
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedB) If X < Y
Explanation:
Solve the equations, x = 3, y = 3.5 - 2x2 – 9x + 10 = 0, 2y2 + 7y – 4 = 0
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or relation cannot be establishedA) If X > Y
Explanation:
2x2 – 9x + 10 = 0
2x2 – 9x – 5x + 10 = 0
Gives x = 2, 5/2
2y2 + 7y – 4 = 0
2y2 + 8y – y – 4 = 0
Gives y = -4, 1/2
put on number line
-4 1/2 2 5/2
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