Question

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

Answer 1

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

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

$\huge\tt{\boxed{\overbrace{\underbrace{\blue{Answer</p><p> }}}}}$

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$\bold{Given}$

$⟹\bold</p><p>{p + q = 13......(i)}$

$⟹</p><p>\bold{pq = 22}$

$⟹</p><p>\bold{p = \frac{22}{q} ......(ii)}$

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

$⟹</p><p>\bold{p + q = 13}$

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

$⟹</p><p>\bold{22 + {q}^{2} = 13q}$

$⟹</p><p>\bold{{q}^{2} - 13q + 22 = 0}$

$⟹</p><p>\bold{{q}^{2} - 11q - 2q + 22 = 0}$

$⟹</p><p>\bold{q(q - 11) - 2(q - 11) = 0}$

$⟹</p><p>\bold{(q - 2)(q - 11) = 0}$

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

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

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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Answer 2

Step-by-step explanation:

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

℘ɧεŋσɱεŋศɭ

☘

\huge\tt\red{\bold{\underline{\underline{❥Question᎓}}}}

❥Question᎓

If p +q =13 and pq = 22, then find value of p2 + q2

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

Answer

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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

\bold{Given}Given

⟹\bold {p + q = 13......(i)}⟹p+q=13......(i)

⟹ \bold{pq = 22}⟹pq=22

⟹ \bold{p = \frac{22}{q} ......(ii)}⟹p=

q

22

......(ii)

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

⟹ \bold{p + q = 13}⟹p+q=13

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

q

22

+q=13

⟹ \bold{22 + {q}^{2} = 13q}⟹22+q

2

=13q

⟹ \bold{{q}^{2} - 13q + 22 = 0}⟹q

2

−13q+22=0

⟹ \bold{{q}^{2} - 11q - 2q + 22 = 0}⟹q

2

−11q−2q+22=0

⟹ \bold{q(q - 11) - 2(q - 11) = 0}⟹q(q−11)−2(q−11)=0

⟹ \bold{(q - 2)(q - 11) = 0}⟹(q−2)(q−11)=0

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

\bold{\blue{Put\: these\: 2 \:values\: of \:q \:in \:equation (ii)}}Putthese2valuesofqinequation(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}}p

2

+q

2

=(11)

2

+(2)

2

=121+4

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

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

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ