Question

if one roots of equation x2 + px + 1 =0 is 2 + root over 3 then values of p and q will be

Answer 1

if one root of a quadratic equation is irrational i.e. in the form of a+sqrt(b) then the other root must be a-sqrt(b).

so the roots of given equation must be 2+sqrt(3) and 2-sqrt(3).

sum of roots=-p=2+sqrt(3)+2-sqrt(3)
=>p=-4

product of roots = q=(2+sqrt(3))(2-sqrt(3))=1

therefore p=-4 & q=1

Answer 2

$\underline\mathfrak{Given:-}$

• $\: \: \: \: \: \: \: P \: {({x})} \: \: = \: \: {x}^{2} \: + \: {px} \: + \: {q}$
• $\: \: \: \: \: \: \: One \: \: root \: \: = \: \: {2} \: + \: \sqrt{3}$

$\underline\mathfrak{To \: \: Find:-}$

• $\: \: \: \: \: value \: \: of \: \: and \: \: q \: .?$

$\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: \: \: If \: \: one \: \: root \: \: is \: \: {2} \: + \: \sqrt{3} \\ \: \: \: \: \: \: \: then, \: \: the \: \: other \: \: root \: \: be \: \: {2} \: - \: \sqrt{3}$

• $\: \: \: \: \: \: \: P \: {({x})} \: \: = \: \: {x}^{2} \: + \: {px} \: + \: {q}$

$\: \: \: \: \: \: \: \therefore\underline{ The \: \: equation \: \: is \: \: in \: \: the \: \: form \: \: of:-}$

$\: \: \: \: \: \: \: \fbox{{ax}^{2} \: + \: {bx} \: + \: {c}}$

• $\: \: \: \: \: \: \: P \: {({x})} \: \: = \: \: {x}^{2} \: + \: {px} \: + \: {q}$

• $\: \: \: \: \: {a} \: \: = \: \: {1}$
• $\: \: \: \: \: {b} \: \: = \: \: {p}$
• $\: \: \: \: \: {c} \: \: = \: \: {q}$

$\: \: \: \: \: \therefore {Sum \: \: of \: \: zeroes} \: \: = \: \: { \: - \: coefficient \: \: of \: \: x}{coefficient \: \: of \: \: {x}^{2}}$

$\: \: \: \: \: \leadsto {({2} \: + \: \sqrt{3})} \: + \: {({2} \: - \: \sqrt{3})} \: \: = \: \: \frac{ \: - \: ({p})}{1}$

$\: \: \: \: \: \leadsto {2} \: + \: {2} \: \: = \: \: \frac{-p}{1}$

$\: \: \: \: \: \leadsto {4} \: \: = \: \: {-p}$

$\: \: \: \: \: \leadsto {p} \: \: = \: \: {-4}$

$\: \: \: \: \: \therefore {Product \: \: of \: \: zeroes} \: \: = \: \: {constant \: \: term}{coefficient \: \: of \: \: {x}^{2}}$

$\: \: \: \: \: \leadsto {({2} \: + \: \sqrt{3})} \: {({2} \: - \: \sqrt{3})} \: \: = \: \: \frac{q}{1}$

$\: \: \: \: \: \leadsto {4} \: + \: {3} \: \: = \: \: \frac{q}{1}$

$\: \: \: \: \: \leadsto {1} \: \: = \: \: {q}$

$\: \: \: \: \: \leadsto {q} \: \: = \: \: {1}$

• $\: \: \: \: \: Hence, \: \: the \: \: value \: \: of \: \: p \: \: and \: \: q \: \: is \: \: {-4} \: \: and \: \: {1}$

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