Question
$2-\sqrt{5},2+\sqrt{5}$
✓
Answer
$\text { Let } \alpha=2-\sqrt{5} \text { and } \beta=2+\sqrt{5}$
$\therefore \alpha+\beta=2-\sqrt{5}+2+\sqrt{5}=4 \text { and } \alpha \beta=(2-\sqrt{5})(2+\sqrt{5})=4-5=1$
$\therefore$ and quadratic equation is, $x^2-(\alpha+\beta) x+\alpha \beta=0$
$\therefore x^2-(4) x+(1)=0 $
$\therefore x^2-4 x+1=0$
$\therefore \alpha+\beta=2-\sqrt{5}+2+\sqrt{5}=4 \text { and } \alpha \beta=(2-\sqrt{5})(2+\sqrt{5})=4-5=1$
$\therefore$ and quadratic equation is, $x^2-(\alpha+\beta) x+\alpha \beta=0$
$\therefore x^2-(4) x+(1)=0 $
$\therefore x^2-4 x+1=0$
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