Question

if the sum of squares of the roots of x square +px-3 is 10 then p=​

Answer 1

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Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$


$\boxed{ \underline{ \bf{Given : }}}$

$p(x) = {x}^{2} +p x - 3$

Sum of the squares of the roots of p(x) = 10.


$\boxed{ \underline{ \bf{To \: \: Find : p }}}$


$\boxed{ \underline{ \bf{ Process \: : }}}$


Let the zeroes of the p(x) be $\alpha \: and \: \beta$


Now,

Sum of the zeroes = $\frac{-b}{a}$

$\implies \: \alpha \: + \: \beta \: = \: \frac{ - b}{a}$

$\implies \alpha \: + \: \beta \: = \: \frac{ - \: p}{1} \: = p \: \: \: \rightarrow(1)$


Product of zeroes = $\frac{c}{a}$

$\implies \: \alpha \times \beta \: = \frac{c}{a}$

$\implies \: \alpha \times \beta \: = \frac{ - \: 3}{1} \: = \: - 3 \: \: \: \rightarrow \: (2)$


Now A/Q,

$\implies \: { \alpha }^{2} \: + \: { \beta }^{2} = \: 10$

$\implies \: {( \alpha \: + \: \beta )}^{2} - 2 \alpha \beta \: = \: 10$

$\implies \: {( - p)}^{2} - \: 2( - \: 3) \: = \: 10$

$\implies \: {p}^{2} \: + \: 6 \: = \: 10$

$\implies \: {p}^{2} = \: 4$

$\boxed{ \bf{ \therefore \: p \: = \: 2 \: \: or \: \: p \: = \: - 2}}$