Question

if p+q+r=16 and pq+qr+rp=25 find the value of p2+q2+r2
​

Answer 1

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(p + q + r)² = p² + q² + r² + 2(pq + qr + rp)

16² = p² + q² + r² + 2(25)

256 = p² + q² + r² + 50

p² + q² + r² = 206

hope it helps..

Answer 2

${\underline{\underline{\frak{\pink{\dag\:Given :-}}}}}$

• p + q + r = 16
• pq + qr + rp = 25

${\underline{\underline{\frak{\pink{\dag\:To\:Find:-}}}}}$

• p² + q² + r²

${\underline{\underline{\frak{\pink{\dag\:Solution:-}}}}}$

We know that

$\star\boxed{ \sf{ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ac) }}$

So, here,

$\colon \rightarrow \sf \: {(p + q + r)}^{2} = {p}^{2} + {q}^{2} + {r}^{2} + 2(pq + qr + rq) \\ \\ \sf \: put \: the \: given \: values \: \\ \\ \colon \rightarrow \sf \: {(16)}^{2} = {p}^{2} + {q}^{2} + {r}^{2} + 2(25) \\ \\ \colon \rightarrow \sf \: 256 = {p}^{2} + {q}^{2} + {r}^{2} + 50 \\ \\ \colon \rightarrow \sf \: {p}^{2} + {q}^{2} + {r}^{2} = 256 - 50 \\ \\ \colon \rightarrow \underline{ \boxed{ \frak{ \red{ {p}^{2} + {q}^{2} + {r}^{2} = \bf 206}}}}$

so,

The value of p² + q² + r² = 206 .

_____________________________

${\underline{\underline{\bf{Some\:Formulas:}}}}$

• (a+b)² = a² + b² + 2ab

• (a-b)² = a² + b² - 2ab

• (a+b+c)² = a² + b² + c² + 2(ab+bc+ac)

• (a+b)³ = a³ + b³ + 3ab(a+b)

• (a-b)³ = a³ - b³ - 3ab(a-b)