If R is such that the sum of the cubes of the roots of x^2+(2-)x+(10-)=0 is minimum, then …

Question

If $\lambda\in\mathbb R$ is such that the sum of the cubes of the roots of $x^2+(2-\lambda)x+(10-\lambda)=0$ is minimum, then the magnitude of the difference of the roots of this equation is

Accepted Answer

Let $$\alpha $$, $$\beta $$ are the roots of the equation,

$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$\lambda $$ - 2 and $$\alpha $$ $\beta $= 10$-$$\lambda $${\alpha ^3} + {\beta ^3}$= ($\alpha $+$\beta $)3$-$3$\alpha $$ $\beta $$ ($$\alpha $$ + $$\beta $$)

= ($$\lambda $$ - 2)3 - 3(10 - $$\lambda $$)($$\lambda $$ - 2)

= $$\lambda ^3$$ - 3$$\lambda ^2$$ - 24$$\lambda $$ + 52

Let $$f(\lambda $$) = $$\lambda ^3$$ - 3$$\lambda ^2$$ - 24$$\lambda $$ + 52

$$ \therefore $$ $${{df(\lambda )} \over {d\lambda }}$$ = 3$$\lambda ^2$$ - 6$$\lambda $$ - 24

$$ \therefore $$ at maximum of minimum $${{df(\lambda )} \over {d\lambda }}$$ = 0

$$ \therefore $$ $$\lambda ^2$$ - 2$$\lambda $$ - 8 = 0

$$ \Rightarrow $$ ($$\lambda $$ + 2) ($$\lambda $$ - 4) = 0

$$ \Rightarrow $$ $$\lambda $$ = -2, 4

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = 2$$\lambda $$ - 2

When $$\lambda $$ = -2

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = - 6 < 0

$$ \therefore $$ at $$\lambda $$ = -2, f($$\lambda $$) has maximum value.

When $$\lambda $$ = 4

$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = 6 > 0

$$ \therefore $$ at $$\lambda $$ = 4, f($$\lambda $$) has minimum value.

$$ \therefore $$ When $$\lambda $$ = 4 equation is,

x2 - 2x + 6 = 0

$$ \therefore $$ ($$\alpha $$ - $$\beta $$)2 = ($$\alpha $$ + $$\beta $$)2 - 4$$\alpha \beta$$

$$ \Rightarrow $$ x2 - 4 $$ \times $$ 6

= - 20

$$ \Rightarrow $$ ($$\alpha $$ - $$\beta $$) = $$2\sqrt 5 i$$

$$ \Rightarrow $$ $$\left| {\alpha  - \beta } \right|$$ = $$2\sqrt 5$$ (ans)

Solutions · 1