Question

If a and b are the zeroes of the polynomial 2x²+5x+k and satisfy a²+b²+ab = 17/4 find the value of k

Answer 1

Answer:

4

Step-by-step explanation:

a, b are the zeroes of the polynomial 2x² + 5k + k

Comparing 2x² + 5x + k with Ax² + Bx + C we get,

• A = 2
• B = 5
• C = k

Sum of zeroes = a + b = - B / A = - 5 / 2

Product of zeroes = ab = C / A = k / 2

Given :

a² + b² + ab = 17/4

Adding ab on both sides

⇒ a² + b² + ab + ab = 17/4 + ab

Substituting ab = k / 2

⇒ a² + b² + 2ab = 17/4 + k/2

Using algebraic identity a² + b² + 2ab = ( a + b )²

⇒ ( a + b )² = ( 17 + 2k ) / 4

⇒ ( - 5 / 2 )² = ( 17 + 2k ) / 4

⇒ 25 / 4 = ( 17 + 2k ) / 4

⇒ 25 = 17 + 2k

⇒ 25 - 17 = 2k

⇒ 8 = 2k

⇒ k = 8 / 2 = 4

Therefore the value of k is 4.