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The minimum value of the sum of the squares of the roots of

$\mathrm{x^{2}+(3-a) x+1=2 a}$ is:
Option: 1
$4$

Option: 2
$5$

Option: 3
$6$

Option: 4
$8$

Answers (1)

best_answer

$\mathrm{x^{2}+(3-a) x+1=2 a} \\$

$\mathrm{x^{2}+(3-a) x+(1-2 a)=0} \\$

$\mathrm{\alpha+\beta=a-3} \\$

$\mathrm{\alpha \beta=1-2 a }.$

$\mathrm{\therefore \quad \alpha^{2}+\beta^{2} =(\alpha+\beta)^{2}-2 \alpha \beta} \\$

$\mathrm{=(a-3)^{2}-2(1-2 a)} \\$

$\mathrm{=a^{2}-6 a+9-2+4 a} \\$

$\mathrm{=a^{2}-2 a+7} \\$

$\mathrm{=(a-1)^{2}+6 }\\$

$\mathrm{\text { Minimum } \text { value }=6}$

Hence correct option is $3$

Posted by

Deependra Verma

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