- #1

the-ever-kid

- 53

- 0

Find the value of [itex]\lambda[/itex] for which the sum of squares of the roots of the equation: [itex]2x^2 + (2\lambda +4)x^2-(1+ \lambda) = 0[/itex] has minimum value.

2. Homework Equations

[itex]\alpha + \beta = -b/a[/itex]

[itex]\alpha\beta = c/a[/itex]

where a,b,c are coefficients of [itex] x^2 , x [/itex] and constant term.

3. The Attempt at a Solution

what i did was figuring out that [itex] \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2(\alpha\beta)[/itex] ... [itex](i)[/itex]

after that i found out [itex]-b/a[/itex] and [itex]c/a[/itex] plugged it in the equation [itex]i[/itex] and got a quadratic in [itex]\lambda[/itex]

[itex]\lambda^2+5\lambda+5[/itex]...[itex](ii)[/itex]

i then differentiated it to get the minima values of [itex]ii[/itex] i got [itex]f(\lambda)=-5/4, \lambda \longrightarrow -5/2[/itex]

**my problem is that the sum of squares of roots cannot be negative, can they?**

and the solution options are [itex]2,3,-3,-2[/itex]

and the solution options are [itex]2,3,-3,-2[/itex]