Question

find the value of a^2+b^2+c^2 if a+b+c=-9 and ab+bc+ac=-15​

Answer 1

Step-by-step explanation:

Since,

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

∴ (-9)² = a² + b² + c² + 2(-15)

81 = a² + b² + c² - 30

∴ a² + b² + c² = 81 + 30 = 111

Answer 2

Given: $a+b+c=-9 \ and\ ab+bc+ac=-15$

We have to find the value of$a^2+b^2+c^2$.

As we know that the formula of identity$(a + b + c)^2$ is:$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

By using the above formula of identity, we are solving the above equation.

We are solving in the following way:

We have,

$a+b+c=-9 \ and\ ab+bc+ac=-15$

Now, from the identity rule:

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

By putting the given values in the above formula:

$=>(-9)^2=a^2+b^2+c^2+2(-15)\\=>81=a^2+b^2+c^2-30\\=>a^2+b^2+c^2=81+30\\=>a^2+b^2+c^2=111$

Hence, the value of$a^2+b^2+c^2\ is\ 111.$