Question

If the sum of zeroes of a quadratic polynomial 3x^2 - kx + 6 is 3 , find the value of k *

Answer 1

$\large{\underline{\bf{\green{Given:-}}}}$

• ✦ sum of zeroes = 3
• ✦ p(x) = 3x² - kx + 6

$\large{\underline{\bf{\green{To\:Find:-}}}}$

• ✦ value of k

$\huge{\underline{\bf{\purple{Solution:-}}}}$

p(x) = 3x² - kx + 6

• a = 3
• b = -k
• c = 6

sum of zeroes = - b/a

➝ 3 = -(-k)/3

➝ 3 = k/3

➝ k = 3×3

➝ k = 9

So,

Value of k is 9.

p(x) = 3x² - kx + 6

➝ 3x² - 9x + 6

➝ 3x² - 3x - 6x +6

➝ 3x(x -1) -6(x -1)

➝ (3x-6) (x -1)

➝ x = 2 or x = 1

Verification :-

sum of zeroes = - b/a

➝ 2 +1 = -(-9)/3

➝ 3 = 3

product of zeroes = c/a

➝ 2×1 = 6/3

➝ 2 = 2

LHS = RHS

hence , Verified

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

Step-by-step explanation:

3x^2-kx + 6=0

x^2-k/3*x + 2

х₁+х₂=-(-к/3) , -(-к/3)=3 , к=9