The values of 'a' for which the sum of yhe squares of the roots of the eqn x²-(a-2)x-a-1=0 assume the least value is-
To find the values of 'a' for which the sum of the squares of the roots of the equation x² - (a-2)x - (a-1) = 0 assumes the least value, we can use the properties of quadratic equations.
Let's solve the equation to find the roots of the equation:
Using the quadratic formula, where the roots are given by:
x = (-b ± √(b² - 4ac)) / 2a
For our equation, a = 1, b = -(a-2) = 2-a, and c = -(a-1) = 1-a.
Substituting these values into the quadratic formula:
x = (-(2-a) ± √((2-a)² - 4(1)(1-a))) / 2(1)
Simplifying further:
x = (a - 2 ± √(4 - 4a + a² - 4 + 4a)) / 2
x = (a - 2 ± √(a² - 2a)) / 2
x = (a - 2 ± √a(a - 2)) / 2
Now, let's find the sum of the squares of the roots. Let the roots be x₁ and x₂.
Sum of the squares of the roots, s = x₁² + x₂²
Substituting the values of x₁ and x₂:
s = [((a - 2 + √a(a - 2)) / 2)²] + [((a - 2 - √a(a - 2)) / 2)²]
simplifying further:
s = [(a - 2 + √a(a - 2))²] + [(a - 2 - √a(a - 2))²]
Expanding and simplifying the equation:
s = (a - 2)² + √a(a - 2))² + 2[(a - 2)(√a(a - 2))]
s = (a - 2)² + a(a - 2) + 2√a²(a - 2)
s = (a - 2)[(a - 2) + a + 2√a²]
s = (a - 2)(2a - 2 + 2√a²)
s = (a - 2)(2a + 2(√a)² - 2)
s = (a - 2)(2a + 2a - 2)
s = (a - 2)(4a - 4)
s = 4a² - 12a + 8
Now, to find the minimum value of s, we can take the derivative of s with respect to 'a' and set it equal to zero:
d(s)/d(a) = 8a - 12
Setting this derivative equal to zero and solving for 'a':
8a - 12 = 0
8a = 12
a = 12/8
a = 1.5
Therefore, the value of 'a' for which the sum of the squares of the roots of the equation assumes the least value is a = 1.5.