Math, asked by itgo5, 8 months ago

subtract 4a-7ab+3b+12 from 12a -9ab +5b -3 ?

Answers

Answered by Anonymous

143

$\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}$

$\huge\tt\red{\bold{\underline{\underline{❥Question᎓}}}}$ If p +q =13 and pq = 22, then find value of p2 + q2

$\huge\tt{\boxed{\overbrace{\underbrace{\blue{Answer }}}}}$

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$\bold{Given}$

$⟹\bold{p + q = 13......(i)}$

$⟹\bold{pq = 22}$

$⟹\bold{p = \frac{22}{q} ......(ii)}$

$\bold{\red{Put\: value\: of\: p \:in \:equation (i)}}$

$⟹\bold{p + q = 13}$

$\bold{ \frac{22}{q} + q = 13}$

$⟹\bold{22 + {q}^{2} = 13q}$

$⟹\bold{{q}^{2} - 13q + 22 = 0}$

$⟹\bold{{q}^{2} - 11q - 2q + 22 = 0}$

$⟹\bold{q(q - 11) - 2(q - 11) = 0}$

$⟹\bold{(q - 2)(q - 11) = 0}$

$\bold{q = 2\: and\ \: q = 11}$

$\bold{\blue{Put\: these\: 2 \:values\: of \:q \:in \:equation (ii)}}$

we get two values of p =(2,11)

Now ,we have find out both values of p and q

$\bold{\pink{ {p}^{2} + {q}^{2} = {(11)}^{2} + {(2)}^{2} = 121 + 4}}$

$\bold{\orange{= 125}}$

Hence ,the value of p^2+q^2=125

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нσρє ıт нєłρs yσυ

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Answered by Anonymous

According to the question

$\bold{\red{Put\: value\: of\: p \:in \:equation (i)}} \\ \\ ⟹ \bold{p + q = 13} \\ \\ \bold{ \frac{22}{q} + q = 13} \\ \\ ⟹ \bold{22 + {q}^{2} = 13q} \\ \\ ⟹ \bold{{q}^{2} - 13q + 22 = 0} \\ \\ ⟹ \bold{{q}^{2} - 11q - 2q + 22 = 0} \\ \\ ⟹ \bold{q(q - 11) - 2(q - 11) = 0} \\ \\ ⟹ \bold{(q - 2)(q - 11) = 0} \\ \\ \bold{q = 2\: and\ \: q = 11} \\ \\ \bold{\blue{Put\: these\: 2 \:values\: of \:q \:in \:equation (ii)}} \\ \\ we \: get \: two \: values \: of \: p \: = \: (2,11) \\ \\ Now , \: we \: have \: find \: out \: both \: values \: of \: p \: and \: q \\ \\ \bold{\pink{ {p}^{2} + {q}^{2} = {(11)}^{2} + {(2)}^{2} = 121 + 4}} \\ \\ \bold{\orange{= 125}}\\ \\Hence , \: the \: value \: of \: p^2+q^2 \: = \: 125$

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