Question

A+b+c=12ab+bc+ac=47a^2+b^2+c^2 -?

Answer 1

Ответ:

Пошаговое объяснение:

3^2+4^2+5^2=9+16+25=50

Answer 2

To find the value of $a^2 + b^2 + c^2$, we need to solve the system of equations:

$$\begin{align*} a + b + c &= 12 \\ ab + bc + ac &= 47 \\ \end{align*}$$

Let's solve this system using substitution. From the first equation, we can express $c$ in terms of $a$ and $b$:

$$c = 12 - a - b$$

Substituting this value of $c$ in the second equation:

$$ab + b(12 - a - b) + a(12 - a - b) = 47$$

Simplifying further:

$$ab + 12b - ab - b^2 + 12a - a^2 - ab = 47$$

Combining like terms:

$$12a + 12b - a^2 - b^2 = 47$$

Rearranging:

$$a^2 + b^2 - 12a - 12b + 47 = 0$$

Now we have a quadratic equation in terms of $a$ and $b$. To solve it, we can complete the square. Let's rewrite the equation:

$$(a^2 - 12a) + (b^2 - 12b) = -47$$

Completing the square for the $a$ terms:

$$(a^2 - 12a + 36) + (b^2 - 12b) = -47 + 36$$

Simplifying further:

$$(a - 6)^2 + (b^2 - 12b) = -11$$

Completing the square for the $b$ terms:

$$(a - 6)^2 + (b^2 - 12b + 36) = -11 + 36$$

Simplifying further:

$$(a - 6)^2 + (b - 6)^2 = 25$$

Now we have a simplified equation that represents a circle in the $ab$-plane. The equation suggests that the sum of the squares of $a$ and $b$ lies on the circle with radius 5 and center at $(6, 6)$.

Unfortunately, without additional information or constraints on the values of $a$, $b$, and $c$, we cannot determine the specific value of $a^2 + b^2 + c^2$ from the given system of equations.