If x² - x = 12 and y² - y = 12, then x - y = ?
- 0
- ±7
- ±1
- ±3
Explanation
Given:
x² − x = 12
y² − y = 12
So both equations are equal:
x² − x = y² − y
⇒ x² − x − y² + y = 0
⇒ (x² − y²) − (x − y) = 0
⇒ (x − y)(x + y − 1) = 0
Now set each factor to zero:
1. x − y = 0 → ⇒ x = y
2. x + y − 1 = 0 → ⇒ x + y = 1
Now plug into original equation to find values:
Try solving x² − x = 12
⇒ x² − x − 12 = 0
⇒ (x − 4)(x + 3) = 0
⇒ x = 4 or x = -3
Case 1: x = 4
If x + y = 1 ⇒ y = -3
Then x − y = 4 − (−3) = 7
Case 2: x = -3
If x + y = 1 ⇒ y = 4
Then x − y = -3 − 4 = −7
So possible values of x − y = 0 or ±7
Related MCQs
- 2/x + 8/(x + 3)
- 4/x + 8/(x + 3)
- 6/x + 8/(x + 3)
- None of these
اس سوال کو وضاحت کے ساتھ پڑھیں
- 4
- 5
- 6
- None of these
اس سوال کو وضاحت کے ساتھ پڑھیں
Leave a Reply
Your email address will not be published. Required fields are marked *
