Question

Используя теорему Виета, найдите подбором корни уравнения:q^2+3q-18=0n^2-12+11=0t^2-11t+18=0 ^ - в квадратеПомогите пожалуйста!!!!(((

Answer 1

1)q*q=-18
   q+q=-3
q=6
q=-3
2)n*n=11
   n+n=12
n=11
n=1
3)t*t=18
  t+t=11
t=9
t=2

Answer 2

Using Vieta's Theorem to Find Roots of Quadratic Equations

To find the roots of the given quadratic equations using Vieta's theorem, we can use the following approach:

For a quadratic equation of the form ax^2 + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a.

Let's solve the given quadratic equations one by one.

1. For the equation q^2 + 3q - 18 = 0: - Here, a = 1, b = 3, and c = -18. - Using Vieta's theorem, the sum of the roots is -b/a = -3/1 = -3 and the product of the roots is c/a = -18/1 = -18. - We need to find two numbers that add up to -3 and multiply to -18. By trial and error, we find that the roots are q = 6 and q = -3.

2. For the equation n^2 - 12n + 11 = 0: - Here, a = 1, b = -12, and c = 11. - Using Vieta's theorem, the sum of the roots is -b/a = 12/1 = 12 and the product of the roots is c/a = 11/1 = 11. - We need to find two numbers that add up to 12 and multiply to 11. By trial and error, we find that the roots are n = 11 and n = 1.

3. For the equation t^2 - 11t + 18 = 0: - Here, a = 1, b = -11, and c = 18. - Using Vieta's theorem, the sum of the roots is -b/a = 11/1 = 11 and the product of the roots is c/a = 18/1 = 18. - We need to find two numbers that add up to 11 and multiply to 18. By trial and error, we find that the roots are t = 9 and t = 2.

So, using Vieta's theorem and trial and error, we have found the roots for the given quadratic equations: - For q^2 + 3q - 18 = 0, the roots are q = 6 and q = -3. - For n^2 - 12n + 11 = 0, the roots are n = 11 and n = 1. - For t^2 - 11t + 18 = 0, the roots are t = 9 and t = 2.

I hope this helps! If you have any more questions or need further assistance, feel free to ask!