Quadratic Equation Quiz for SBI PO Pre, IBPS PO Pre, SBI Clerk, IBPS Clerk, 2018

Join Testzone, Smartkeeda’s Test Series Platform for:

Directions: In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer.
» Explain it
A
I.  21  +  11  = 7x
x x

21 + 11  = 7x
x

32  = 7x
x

x =  32
7

II. 2y2 – 11y + 12 = 0

2y2 – 8y – 3y + 12 = 0

2y (y – 4) – 3 (y – 4) = 0

(2y – 3) (y – 4) = 0

y =  3  , 4
2

x > y

Hence, option A is correct.

» Explain it
C
I. 143x2 + 43x – 66 = 0

or, 143x2 + (121 – 78)x – 66 = 0

or, 143x2 + 121x – 78x – 66 = 0

or, 11x (13x + 11) – 6 (13x + 11) = 0

or, (13x + 11) (11x – 6) = 0

x = 6/11, - 11 /13
 
II. 33y2 – 40y + 12 = 0

or, 33y2 – (22 + 18)y + 12 = 0

or, 33y2 – 22y – 18y + 12 = 0

or, 11y (3y – 2) – 6 (3y – 2) = 0

or, (11y – 6) (3y – 2) = 0

y = 6/11, 2/3
 
x ≤ y

Hence, option C is correct.
 
3
I. 72x2 – 101x + 35 = 0             II. 45y2 – 62y + 21 = 0
» Explain it
C
I. 72x2 – 101x + 35 = 0

or, 72x2 – (56 + 45)x + 35 = 0

or, 72x2 – 56x – 45x + 35 = 0

or, 8x (9x – 7) – 5 (9x – 7) = 0

(8x – 5) (9x – 7) = 0

x = 5/8, 7/9
 
II. 45y2 – 62y + 21 = 0

or, 45y2 – (35 + 27)y + 21 = 0

or, 45y2 – 35y – 27y + 21 = 0

or, 5y (9y – 7) – 3 (9y – 7) = 0

(5y – 3) (9y – 7) = 0

y = 3/5, 7/9
 
While comparing the values of x and y, one root value of x lies between the two root values of y

Hence, option C is correct.

4
» Explain it
A
I. 22x2 + 611x – 8 = 0

or, 22x2 + (811 – 211)x – 8 = 0

or, 22x2 + 811x – 211x – 8 = 0

or, 211x (11x + 4) – 2(11x + 4) = 0

(211x – 2) (11x + 4) = 0

x = 1/11, - 4/11
 
II. 3y2 + 14y – 5 = 0

or, 3y2 + (15 – 1)y – 5 = 0

or, 3y2 + 15y – y – 5 = 0

or, 3y (y + 5) – 1 (y + 5) = 0

(3y – 1) (y + 5) = 0

y = 1/3, -5
 
While comparing the values of x and y, one root value of y lies between the root values of x

Hence, option A is correct.
 
5
I. 56x2 + 15x – 56 = 0              II. 7y2 – 34y – 48 = 0
» Explain it
B
I. 56x2 + 15x – 56 = 0

or, 56x2 + (64 – 49)x – 56 = 0

or, 56x2 + 64x – 49x – 56 = 0

or, 8x (7x + 8) – 7 (7x + 8) = 0

(7x + 8) (8x – 7) = 0

x = 7/8, -8/7
 
II. 7y2 – 34y – 48 = 0

or, 7y2 – (42 – 8)y – 48 = 0

or, 7y2 – 42y + 8y – 48 = 0

or, 7y (y – 6) + 8 (y – 6) = 0

(7y + 8) (y – 6) = 0

y = -8/7, 6
 
While comparing the values of x and y, one root value of y lies between the root values of x

Hence, option B is correct.