Question

If a+b+c = 9 and ab + bc + ac = 26. Find a^2 + b^2 + c^2​

Answer 1

Answer:

(a+b+c)² = a² + b² + c² + 2(ab+bc+ca)

Given:

a+b+c = 9 & ab+bc+ca = 26

Now putting the values in the above formula,

We get,

(9)² = a² + b² + c² + 2×26

81 = (a² +b² +c²) + 52

(a² +b²+c²) = 81-52

(a² + b² + c²) = 29 (Answer)

Answer 2

$\large \green {\colorbox{cyan}{ \colorbox{black}{answer}}}$

29

$\large \green {\colorbox{cyan}{ \colorbox{black}{explanation}}}$

$\sf Given \: a + b + c = 9 \: and \: ab + bc + ac = 26$

$\sf we \: know \: {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2}$

$\sf + \: 2ab \: + 2bc \: + 2ca$

$\sf⇒ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(a + b + c)$

putting the values here we get

$\sf {(9)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2 \: (26)$

$\sf⇒81 = {a}^{2} + {b}^{2} + {c}^{2} + 52$

$\sf⇒ {a}^{2} + {b}^{2} + {c}^{2} = 81 - 52 = 29$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\large \boxed {hope \: it \: helps}$

:)