In the following question, two equations are given. On their basis you have to determine the relation between x and y and then give answer I. 3x^2 – 7x + 2 = 0 II. 2y^2

A. x < y
B. x > y
C. x ≤ y
D. x ≥ y

Answer

A. x < y

Solution

Equation I
3x² – 7x + 2 = 0
3x²-6x-x+2=0
3x (x-2)-1(x-2)=0
(3x -1)(x – 2) = 0

3x -1 = 0
3x =1
x = 1/3 , x-2=0
x = 2

Equation II
2y² -11y +15 = 0
2y² – 6y – 5y +15 = 0
2y (y-3)-5(y-3)=0
(2y- 5)(y-3)= 0
(2y-5)=0
y = 5/2
Also (y – 3) = 0
y=3

Now comparing x and y

x = 1/3 is smaller than 5/2 and 3
x= 2 is smaller than 3 and 5/2

Thus we see that x is smaller than y

x<y

Post Views: 281