Question

if a+b+c =10 and a² +b²+c²=38 find ab+bc+ca​

Answer 1

$\color{Red}{Answer}$

a + b + c = 10

⇒ (a + b + c)2 = (10)2

⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 100

⇒ 38 + 2(ab + bc + ca) = 100

⇒ 2(ab + bc + ca) = 62

⇒ 2(ab + bc + ca) = 62

⇒ (ab + bc + ca) = 62/2

⇒ ab + bc + ca = 31

Answer 2

$\huge \sf \underline{\underline{\pink{Answer}}} :$

$\large \sf ab+bc+ca = 31$

$\huge \sf\underline{\underline{\pink{Solution}}} :$

$\large \sf \color{red}{\underline{{Given}}} :$

✯ a + b + c = 10

✯ a² + b² + c² = 38

we know ,

$\large \sf \color{green}\boxed{\sf (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca}$

➣ a + b + c = 10

➣ (a+b+c)² = 10²

$\implies \sf a^2+b^2+c^2+2ab+2bc+2ca = 100 \\ \\ \sf \implies (a^2+b^2+c^2) + 2(ab+bc+ca) = 100 \\ \\ \implies \sf 38 + 2(ab+bc+ca) = 100 \\ \\ \implies \sf 2(ab+bc+ca) = 62 \\ \\ \implies \sf ab+bc+ca = \frac{\cancel{62}}{\cancel{2}} \\ \\ \implies \sf \boxed{\sf ab+bc+ca = 31}$