Question

Q) The Value of a for which the sum of squares of the roots of the equation
${x}^{2} - (a - 2)x - a - 1 = 0$
assume the least value
​

Answer 1

Answer :-) 1

Solution :-)

$\sf Let \: \alpha \: and \: \beta \: be \: the \\ \sf roots \: of \: equation \\ \sf \text {x}^{2} - \text (a - 2) \text x - a - 1 = 0 \\ \\ \sf Then \\ \\ \sf Sum \: of \: roots \: = \alpha + \beta = a - 2 \\ \\ \sf Product \: of \: roots = \alpha \beta - a - 1 \\ \\ \sf \: as \\ \\ \sf { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta ) }^{2} - 2 \alpha \beta \\ \\ \implies { \alpha }^{2} + { \beta }^{2} = (a - 2) {}^{2} + 2(a + 1) \\ \\ \implies { \alpha }^{2} + { \beta }^{2} = {a}^{2} - 2a + 6 \\ \\ \implies { \alpha }^{2} + { \beta }^{2} =(a - 1) {}^{2} + 5 \\ \\ \sf So \: the \: value \: of \: { \alpha }^{2} + { \beta }^{2} \: will \: be \: least \\ \\ \sf if \: a - 1 = 0 \\ \\ \huge \implies \bold \red{ \boxed{ \boxed{a = 1}}}$

Alternate Solution :-)

$\sf As \: \: \: \alpha + \beta = a - 2 \\ \\ \sf and \: \: \: \alpha \beta = - a - 1 \\ \\ \sf Let \\ \sf f(a) = { \alpha }^{2} + { \beta }^{2} = ( \alpha + \beta ) {}^{2} - 2 \alpha \beta \\ \\ = (a - 2) {}^{2} + 2(a + 1) = {a}^{2} - 2a + 6 \\ \\ \implies \sf f '(a) = 2a - 2$

For Maxima and Manima Put f'(a) = 0

$2a - 2 = 0 \\ \\ a = 1 \\ \\ \sf now \: \: f''(a) = 2 \\ \\ \implies \sf f''(1) = 2 > 0$

So f(a) is minimum at a = 1

Answer 2

$\qquad \qquad \underline{\pmb{\mathbb{\bigstar \:\:\:POLYNOMIAL \:\::}{\: x^2 \:-\: ( \:a - 2 )x\: - a - 1 \: \:\:}}}$

As , We know that ,

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Sum \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha + \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\} \: \:=\:\dfrac{ - \:( Cofficient \:of \: x \:)\:}{Cofficient \:of \:x^2 \:} \\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\}\:\:=\:\:\:\:\dfrac{ - \:( Cofficient \:of \: x \:)\:}{Cofficient \:of \:x^2 \:}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:+\:\beta \:\Big\}\:\:=\:\:\:\:\dfrac{ - \:\big[ -\{ a - 2 \:\}\big]\:}{1 \:}\\\\\dashrightarrow \sf \: \alpha \:+\:\beta \: \:\:=\:\:\dfrac{ ( a - 2 )}{1}\\\\\dashrightarrow \sf \: \alpha \:+\:\beta \: \:\:=\:\:a - 2 \:\\\\\dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \alpha \:+\:\beta \: \:\:=\:\:a - 2 \:\:}}}}}\:\:\bigstar$

⠀⠀⠀⠀⠀AND ,

$\qquad \underline {\boxed {\pmb{ \:\maltese \:Product \:\: of \:\:zeroes \:\:\purple{ \:(\: \alpha \beta \;)\:} \:: \: }}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\} \: \:=\:\dfrac{ \:Constant \:Term\:}{Cofficient \:of \:x^2 \:} \\\\\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\}\:\:=\:\:\:\:\dfrac{ Constant \:Term\:}{Cofficient \:of \:x^2 \:}\\\\\dashrightarrow \sf \Big\{ \: \alpha \:\:\beta \:\Big\}\:\:=\:\:\:\:\dfrac{ -( a - 1 )\:}{1 \:}\\\\\dashrightarrow \sf \: \alpha \:\:\beta \: \:\:=\:\:\dfrac{-( a - 1 )}{1}\\\\\dashrightarrow \sf \: \alpha \:+\:\beta \: \:\:=\:\:-( a - 1) \:\\\\ \dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: \alpha \:\:\beta \: \:\:=\:\: a + 1 \:\:}}}}}\:\:\bigstar$

━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀¤ Finding Minimum value of a for sum of squares of the roots ( α²+β² ) :

$\qquad \dashrightarrow \sf \alpha ^2 + \beta^2 \:$

As , we know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ Algebraic \:Indentity \:\:: a^2 + b^2 =\:( a + b )^2 - 2ab }\bigg\rgroup \\\\\qquad \dashrightarrow \sf \alpha ^2 + \beta^2 \:\\\\\qquad \dashrightarrow \sf \alpha ^2 + \beta^2 \:=\: ( \alpha + \beta )^2 - 2 \alpha \beta \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf \alpha ^2 + \beta^2 \:=\: ( \alpha + \beta )^2 - 2 \alpha \beta \:$

$\qquad \dashrightarrow \sf \alpha ^2 + \beta^2 \:=\: ( a - 2 )^2 - 2 ( a + 1 )\:\\\\\qquad \dashrightarrow \sf \alpha ^2 + \beta^2 \:=\: a^2 - 2a + 6\:\\\\\qquad \dashrightarrow \sf \alpha ^2 + \beta^2 \:=\: ( a -1 )^2 + 5 \:$

⠀⠀⠀⠀⠀We have to find minimum value :

Therefore ,

$\qquad \dashrightarrow \sf \alpha ^2 + \beta^2 \:=\: ( a -1 ) \:\\\\\qquad \dashrightarrow \sf ( a - 1 ) \:\: \:\\\\\qquad \dashrightarrow \sf a \:=\: 1 \:\\\\\dashrightarrow \underline {\boxed {\pmb{\pink{ \frak { \: a\: \:\:=\:\:1 \:\:}}}}}\:\:\bigstar$

$\qquad\therefore \underline {\sf Hence, \:The \:Value \:of \:a \:is \:\pmb{\bf 1 }\:\:}$