Question
The roots of x² − (k+3)x + (3k − 1) = 0 are real
and distinct, and the larger root exceeds the smaller by 5. Find k.Solution
ATQ,
Let roots a, b with a − b = 5 and a + b = k+3, ab = 3k − 1. Then (a − b)² = (a + b)² − 4ab ⇒ 25 = (k+3)² − 4(3k − 1). ⇒ 25 = k² + 6k + 9 − 12k + 4 ⇒ 0 = k² − 6k − 12. k = [6 ± √(36 + 48)]/2 = [6 ± √84]/2 = [6 ± 2√21]/2 = 3 ± √21. Both make distinct real roots. Answer: k = 3 ± √21.
- What will come in the place of question mark (?) in the given expression?
(198/13) X (52/11) - ? ÷ 5 = 13 + 68 ÷ 4 What will come in the place of question mark (?) in the given expression?
√4096 + √3249 = (?)2
Simplify the following expressions and choose the correct option.
(96 ÷ 8 + 72 ÷ 9) * 3
((12+12+12+12)÷4)/((8+8+8+8+8+8)÷16) = ?
{(5/8) + (4/5)} × (?/19) = 33
(72 × 52 + 1555 )/(79+60) = 2000 ÷ ?
What will come in the place of question mark (?) in the given expression?
25% of 1280 + (41 × 4) = ?2
- What will come in place of (?) in the given expression.
(14)² – (12)² = ? What value should come in the place of (?) in the following questions?
√(60 + 82 + 101) * 5 = ?
