Advertisements
Advertisements
Question
Find the value of k such that the polynomial x2-(k +6)x+ 2(2k - 1) has some of its zeros equal to half of their product.
Advertisements
Solution
For given polynomial 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(2k-1))/1`
⇒ k + 6=2k-1
⇒ 6+1= 2k - k
⇒ k = 7
Therefore, the value of k = 7.
APPEARS IN
RELATED QUESTIONS
The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x).
If 1is a zero of the quadratic polynomial `ax^2 – 3(a – 1)x – 1`is 1, then find the value of a.
Find the zeroes of the quadratic polynomial `f(x) = 4sqrt3x^2 + 5x – 2sqrt3`.
If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it ______.
If x3 + 1 is divided by x2 + 5, then the possible degree of quotient is ______.
If p(x) is a polynomial of at least degree one and p(k) = 0, then k is known as ______.
A polynomial of degree n has ______.
If the graph of a polynomial intersects the x-axis at only one point, it cannot be a quadratic polynomial.
The graph of y = f(x) is shown in the figure for some polynomial f(x).
The number of zeroes of f(x) is ______.
The given linear polynomial y = f(x) has
