Advertisements
Advertisements
Question
Find the value of k such that the polynomial x2 − (k + 6)x + 2(2k −1) has sum of its zeros equal to half of their product.
Advertisements
Solution
Given polynomial is x2 - (k + 6) x + 2 (2k – 1)
Here
a = 1, b = - (k + 6), c = 2 (2k – 1)
Given that,
Sum of zeroes = `(1)/(2)` product of zeroes
⇒ `[-[-("k"+6)]]/(1) = (1)/(2) xx (2(2"k" - 1))/(1)`
⇒ `"k" + 6 = 2"k" - 1`
⇒ `6 + 1 = 2"k" - "k"`
⇒ `"k" = 7`
APPEARS IN
RELATED QUESTIONS
Find the zeroes of the quadratic polynomial `f(x) = x^2 + 3x ˗ 10` and verify the relation between its zeroes and coefficients.
If one zero of the polynomial `x^2-4x+1 is (2+sqrt3)` , write the other zero.
If 1is a zero of the quadratic polynomial `ax^2 – 3(a – 1)x – 1`is 1, then find the value of a.
If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is -1, then the product of the
other two zeroes is ______.
A quadratic polynomial, whose zeores are -4 and -5, is ______.
The zeroes of the quadratic polynomial x² + 1750x + 175000 are ______.
If one of the zeroes of the quadratic polynomial (k -1)x² + kx + 1 the value of k is ______.
If f(x) = 5x - 10 is divided by x – `sqrt2`, then the remainder will be ______.
If the zeroes of the quadratic polynomial ax2 + bx + c, c ≠ 0 are equal, then ______.
The number of polynomials having zeroes – 3 and 4 is ______.
