If the difference of the zeroes of the quadratic polynomial x2 + kx +45 is equal lo 144, then the value of k is (a) +9 (b) 12 (c) +15 (d) +18
Answers
Correct Question :- If the square of the difference of the zeroes of the quadratic polynomial x² + kx +45 is equal to 144, then the value of k is (a) ±9 (b) ±12 (c) ±15 (d) ±18
Concept used :- for a quadratic polynomial ax² + bx + c :-
- Sum of zeros = - (coefficient of x) /(coefficient of x²) = (-b/a)
- Product Of zeros = Constant Term / (coefficient of x²) = c/a
Solution :-
Let us assume that, the zeroes of the quadratic polynomial x² + kx + 45 are p and q where p > q .
So,
→ sum of zeroes = p + q = (-b/a) = (-k/1) = (-k) ----- Eqn.(1)
→ product of zeroes = pq = (c/a) = 45/1 = 45 ------ Eqn.(2)
also,
→ (p - q)² = 144 (given)
→ p² + q² - 2pq = 144
adding and subtracting 2pq in LHS,
→ p² + q² + 2pq - 2pq - 2pq = 144
→ (p + q)² - 4pq = 144
putting values from Eqn.(1) and Eqn.(2) now,
→ (-k)² - 4 * 45 = 144
→ k² - 180 = 144
→ k² = 144 + 180
→ k² = 324
square root both sides,
→ k = ± 18 (d) (Ans.)
Hence, the value of k is equal to 18 and (-18) .
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