Question

ПОМОГИТЕ, ПРОШУ!!!Докажите, что при любых значениях a,b и с многочлен:а) a^2+9b^2+c^2-6ab-2ac+6bc принимает неотрицательные значения;б) a^2+4b^2-4ab-10a+20b+26 принимает положительные значения.РЕШИТЕ С ПОДРОБНЫМ РЕШЕНИЕМ.

Answer 1

б)

=a^2+4b^2-4ab-10a+20b+26=

=a^2-4ab+4b^2      -10(a+2b)   +25  +1=

=(a-2b)^2  -2*5*(a-2b) +5^2  +1=

=(a-2b-5)^2  +1

так как  (a-2b-5)^2  >= 0 следовательно 

(a-2b-5)^2  +1  принимает положительные значения при любых значениях a,b

Answer 2

а) a^2+9b^2+c^2-6ab-2ac+6bc принимает неотрицательные значения;

a2+9b2+c2-6ab-2ac+6bc = (a-3b-c)2  ; Это выражение будет  принимать неотрицательные значения при любых a, b и c, т.к является квадратом (при возведении в чётную степень любое число будет иметь неотрицательное значение)

Подробности: Применил формулу квадрата суммы (a+b+с)^2=a^2+b^2+2ab+2bc+2ac, в данном случае   (a-3b-c)^2=a^2+(-3b)^2+(-c)^2+(2*a*(-3b))+(2*a*(-c))+2*(-3b)*(-c)=a^2+9b^2+c^2-6ab-2ac+6bc. Я всего лишь обратил процесс.

Answer 3

Proof for Polynomial (a)

To prove that the polynomial a^2 + 9b^2 + c^2 - 6ab - 2ac + 6bc takes non-negative values for any values of a, b, and c, we need to show that the polynomial is always greater than or equal to zero.

Let's consider the polynomial as a quadratic function in terms of variable 'a'. To find the minimum value of this quadratic function, we can use the discriminant.

The discriminant (D) of a quadratic function ax^2 + bx + c is given by the formula: D = b^2 - 4ac.

In our case, the quadratic function is a^2 - (6b + 2c)a + (9b^2 + c^2 + 6bc).

The discriminant (D) of this quadratic function is: D = (6b + 2c)^2 - 4(9b^2 + c^2 + 6bc).

To prove that the polynomial takes non-negative values, we need to show that the discriminant is less than or equal to zero.

Expanding the discriminant, we get: D = 36b^2 + 24bc + 4c^2 - 36b^2 - 4c^2 - 24bc = 0.

Since the discriminant is zero, the quadratic function has a minimum value of zero. Therefore, the polynomial a^2 + 9b^2 + c^2 - 6ab - 2ac + 6bc takes non-negative values for any values of a, b, and c.

Proof for Polynomial (b)

To prove that the polynomial a^2 + 4b^2 - 4ab - 10a + 20b + 26 takes positive values for any values of a and b, we need to show that the polynomial is always greater than zero.

Let's consider the polynomial as a quadratic function in terms of variable 'a'. To find the minimum value of this quadratic function, we can use the discriminant.

The discriminant (D) of a quadratic function ax^2 + bx + c is given by the formula: D = b^2 - 4ac.

In our case, the quadratic function is a^2 - (4b + 10)a + (4b^2 + 20b + 26).

The discriminant (D) of this quadratic function is: D = (4b + 10)^2 - 4(4b^2 + 20b + 26).

To prove that the polynomial takes positive values, we need to show that the discriminant is greater than zero.

Expanding the discriminant, we get: D = 16b^2 + 80b + 100 - 16b^2 - 80b - 104 = 0.

Since the discriminant is zero, the quadratic function has a minimum value of zero. Therefore, the polynomial a^2 + 4b^2 - 4ab - 10a + 20b + 26 takes positive values for any values of a and b.

I hope this helps! Let me know if you have any further questions.