1. Find the value of 'k' such that the quadratic polynomial x2 - (k+6)x+ 2 (2k +1)
has sum of the zeros is half of their product.
Answers
Answer:
k = 5
Step-by-step explanation:
➤ Quadratic Polynomials :
✯ It is a polynomial of degree 2
✯ General form :
ax² + bx + c = 0
✯ Determinant, D = b² - 4ac
✯ Based on the value of Determinant, we can define the nature of roots.
D > 0 ; real and unequal roots
D = 0 ; real and equal roots
D < 0 ; no real roots i.e., imaginary
✯ Relationship between zeroes and coefficients :
✩ Sum of zeroes = -b/a
✩ Product of zeroes = c/a
________________________________
Given quadratic polynomial,
x² - (k + 6)x + 2(2k + 1)
It is of the form ax² + bx + c
a = 1 , b = -(k + 6) , c = 2(2k + 1)
we know,
⇒ Sum of zeroes = -b/a
⇒ Product of zeroes = c/a
According to the question,
Sum of the zeroes = 1/2 (product of the zeroes)
The value of k is 5.
Verification :
⇒ Sum of zeroes = -b/a
= -[-(k + 6)]/1
= k + 6
= 5 + 6
= 11
⇒ Product of zeroes = c/a
= 2(2k + 1)/1
= 4k + 2
= 4(5) + 2
= 20 + 2
= 22
Sum of zeroes = half of the product of zeroes
11 = 1/2 × (22)
11 = 22/2
11 = 11
Hence verified!
Answer:
Step-by-step explanation:
Let α and β are the roots of given quadratic equation x² - ( k +6)x + 2(2k +1) = 0 [ you did a mistake in typing the equation, I just correct it ]
Now, sum of roots = α + β = - {-( k + 6)}/1 = (k + 6)
product of roots = αβ = 2(2k + 1)/1= 2(2k + 1)
A/C to question,
The sum of roots ( zeros ) = 1/2 × products of roots zeros
⇒ (k + 6) = 1/2 × 2(2k + 1)
⇒ (k + 6) = (2k + 1)
⇒ k + 6 = 2k + 1
⇒ k = 5
Hence, k = 5