If alpha and beta are zeros of a polynomial f(x)=x squared -6x+k find value of k such that alpha squared + beta squared =40
let's call the roots a and b, just for ease of typing.
f(x) = x^2-6x+k
recall the sum and product of the roots:
a+b = 6
ab = k
Now, we have
a^2 + b^2 = 40
but,
a^2 + b^2 = (a+b)^2 - 2ab
So,
36 - 2k = 40
To find the value of k such that α^2 + β^2 = 40, we need to use the relationship between the roots of a quadratic polynomial and its coefficients.
Let's start with the quadratic polynomial f(x) = x^2 - 6x + k. The sum of the roots of a quadratic equation is equal to the negative coefficient of the linear term divided by the coefficient of the quadratic term. In this case, the sum of the roots is 6.
Now, we need to find the relationship between the sum of the squares of the roots and the coefficients. Let's consider the squares of the roots α and β:
α^2 + β^2 = (α + β)^2 - 2αβ
From Vieta's formulas, we know that α + β = 6 and αβ = k (product of roots).
Substituting these values into the equation, we get:
α^2 + β^2 = (6)^2 - 2k
α^2 + β^2 = 36 - 2k
We want the sum of the squares of the roots to be 40. Therefore, we set up the equation:
36 - 2k = 40
Solving for k:
-2k = 40 - 36
-2k = 4
k = -2
So, the value of k that satisfies α^2 + β^2 = 40 is k = -2.