Question

if a+b+c=9 and ab+bc+ca=40 find a²+b²+c²

Answer 1

$(a + b + c) {}^{2} = a {}^{2} + b {}^{2} + c {}^{2} + 2(ab + bc + ca)$

given that a+b+c = 9 and ab+bc+ca = 40

$(81) = a {}^{2} + b {}^{2} + c {}^{2} + 2(40)$

$81 - 80 = a {}^{2} + b {}^{2} + c {}^{2}$

$a {}^{2} + b {}^{2} + c {}^{2} = 1$

mark me as brainliest

Answer 2

Step-by-step explanation:

Given, a + b + c = 9 and ab + bc + ca = 40

We know that,

(a + b + c)2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca

⇒ a 2 + b 2 + c 2 = (a + b + c)2 – 2 (ab + bc + ca)

⇒ a 2 + b 2 + c 2 = (9)2 – 2 × 40 = 81 – 80 = 1 [a + b + c = 9 and ab + bc + ca = 40]

Thus, the value of a 2 + b 2 + c2 is 1

Plz mark me as the brainliest and I hope it helps