Let $p,q\; \epsilon\; \textbf{R}.$ If $2-\sqrt{3}$ is a root of the quadratic equation,$x^{2}+px+q=0,$ then : Option 1)  $p^{2}-4q+12=0$           Option 2)  $q^{2}-4p-16=0$ Option 3) $q^{2}+4p+14=0$ Option 4) $p^{2}-4q-12=0$

S solutionqc

Given that one root is $2-\sqrt{3}$

then another root will by $2+\sqrt{3}$

Sum of roots = $\alpha +\beta =4=-p$

Product of roots= $2\beta =1=q$

$\therefore p^{2}-4q-12=0$

Question is wrong

Option 1)

$p^{2}-4q+12=0$

Option 2)

$q^{2}-4p-16=0$

Option 3)

$q^{2}+4p+14=0$

Option 4)

$p^{2}-4q-12=0$

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