If $\alpha$ and $\beta$ are roots of the equation $x^2+p x+\frac{3 p}{4}=0$, such that…
If $\alpha$ and $\beta$ are roots of the equation $x^2+p x+\frac{3 p}{4}=0$, such that $|\alpha-\beta|=\sqrt{10}$, then $p$ belongs to the set :
$\{2,-5\}$
$\{-3,2\}$
$\{-2,5\}$
$\{3,-5\}$
Solution
Let's solve this step by step:
(1) We start with a quadratic equation where $\alpha$ and $\beta$ are its roots:
$x^2+(2-\lambda)x+(10-\lambda)=0$
(2) Using the quadratic formula, we can express these roots as:
$\frac{\lambda-2 \pm \sqrt{4-4\lambda+\lambda^2-40+4\lambda}}{2}$
(3) This can be simplified to:
$\frac{\lambda-2 \pm \sqrt{\lambda^2-36}}{2}$
(4) The magnitude of the difference between these roots is:
$|\alpha - \beta| = |\sqrt{\lambda^2-36}|$
(5) For the sum of cubes of the roots ($\alpha^3 + \beta^3$), we get:
$\alpha^3 + \beta^3 = \frac{(\lambda-2)^3}{4} + \frac{3(\lambda-2)(\lambda^2-36)}{4}$
(6) This expression can be further simplified:
$\frac{(\lambda-2)(4\lambda^2-4\lambda-104)}{4}$
$= (\lambda-2)(\lambda^2-\lambda-26)$
$= f(\lambda)$
(7) Finally, we can observe that this function $f(\lambda)$ attains its minimum value when $\lambda = 4$.
Therefore, when $\lambda = 4$, the function reaches its minimum value.