## quadratic equation quiz

# quadratic equation quiz

Direction (1-10) : In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer.

If

(1) x > y

(2) x y

(3) x < y

(4) x y

(5) x = y or the relationship cannot be established

- I. 8x2 – 3x = 38
- 6y2 + 34 = 39y

- I. 7x2 + 15x – 18 = 0
- 2y2 – 13y + 21 = 0

- I. 3x2 – 15x + 18 = 0
- y2 + 13y = –42

- I. 2x + 3y = 13
- 4x + y = 6

- I. x2 = 529
- y2 + 241 = 770

- I. 2x^2 + 11x + 14 = 0 and
- 4y^2 + 12y +9 =0

- I. x^2-7x+12=0 and
- y^2+y-12=0

- I. x^4- 227= 398 and
- y^2 + 321=346

- I. x^2-1=0 and
- y^2+4y+3=0

- I. x^2-7x+12=0 and
- y^2-12y+32=0

**Answers and Solution**

- (5)

8x2 – 3y – 38 = 0

X = 198, -2

6y2 – 29y + 34 = 0

Y = 176, 2

- (3)
- 7x2 + 15x – 18 = 0

7x2 + 21x – 6x – 18 = 0

7x(x + 3) – 6(x + 3) = 0

(7x – 6) (x + 3) = 0

X = 67, -3

- 2y2 – 13y + 21 = 0

2y2 – 6y – 7y + 21 = 0

2y(y – 3) – 7(y – 3) = 0

(2y – 7) (y – 3) = 0

Y = 73, 3

- (1)
- 3x2 – 15x + 18 = 0

x2 – 5x + 6 = 0

x2 – 2x – 3x + 6 = 0

x(x – 2) – 3(x – 2) = 0

(x – 3) (x – 2) = 0

x = 3, 2

- y2 + 13y = – 42

y2 + 13y + 42 = 0

y2 + 7y + 6y + 42 = 0

y(y + 7) + 6(y + 7) = 0

(y + 6) (y + 7) = 0

y = – 6, –7

- (3)

2x + 3y = 13 ………. (i)

4x + y = 6 ………. (ii)

Now, equation (i) × 2 – equation (ii),

4x + 6y – 4x – y = 26 – 6

5y = 20

y = 4

Put the value of y in equation (ii),

4x + 4 = 6

4x = 2

x = 12

- (5) x2 = 529

x = +23, –23

- y2 + 241 = 770

y2 = 770 – 241

y2 = 529

y = + 23, – 23

- (3)

X = -8/2 , -3/2

Y = -6/4 , -6/4

-3/2 , -3/2

- (2)

X = 3, 4

Y = -4 , 3

- (5)

X = -5 , +5

Y = -5 , +5

- (2)

X = -1 , +1

Y = -3 , -1

- (4)

X = 3 , 4

Y = 4, 8