Question

if p and q are the roots ofbthe equation x^2 - px + q = 0 then
a) p = 1 , q = -2
b) p = 0 , q= 1
c) p = -2 , q = 0
d) p= -2 , q =1​

Answer 1

Answer:

I think the answer is option c

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:p \: and \: q \: are \: roots \: of \: {x}^{2} - px + q = 0$

We know that,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:pq = \dfrac{q}{1}$

$\bf\implies \:pq = q$

$\bf\implies \:\boxed{ \tt{ \: p = 1 \: }}$

Also,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\bf\implies \:p + q = - \dfrac{( - p)}{1}$

$\bf\implies \:p + q =p$

$\bf\implies \:q =0$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \bf\implies \:\begin{cases} &\sf{p = 1} \\ \\ &\sf{q = 0} \end{cases}\end{gathered}\end{gathered}$

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Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d = 0, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$