Question

Сумма корней уравнения 5(m-1)(m-3)(m-2)(m-4)-120=0

Answer 1

5(m-1)(m-3)(m-2)(m-4)-120=0

(5m-5)(m-3)(m-2)(m-4)-120=0

(5m²-15m-5m+15)×(m-2)(m-4)-120=0

(5m³-10m²-20m²+40m+15m-30)×(m-4)-120=0

(5m³-30m²+55m-30)×(m-4)-120=0

(5m⁴-20m³-30m³+120m²+55m²-220m-30m+120-120=0

5m⁴-50m³+175m²-250m=0

5m×(m³-10m²+35m-50)=0

5m×(m³-5m²-5m²+25m+10m-50)=0

5m×(m²×(m-5)-5m×(m-5)+10×(m-5))=0

5m×(m-5)×(m²-5m+10)=0 | :5

m×(m-5)×(m²-5m+10)=0

m=0 или m-5=0 или m²-5m+10=0

m=5. m≠R

m1=0 и m2=5

Ответ: m1=0 ; m2=5

Answer 2

To find the sum of the roots of the given equation, we can start by simplifying the equation and then applying Vieta's formulas.

The given equation is: 5(m-1)(m-3)(m-2)(m-4) - 120 = 0

Step 1: Simplify the equation

Expand the product of the factors: 5(m-1)(m-3)(m-2)(m-4) = 5(m^4 - 10m^3 + 33m^2 - 36m + 24)

Now, the equation becomes: 5(m^4 - 10m^3 + 33m^2 - 36m + 24) - 120 = 0

Step 2: Bring all terms to one side of the equation: 5m^4 - 50m^3 + 165m^2 - 180m + 120 - 120 = 0

Simplify further: 5m^4 - 50m^3 + 165m^2 - 180m = 0

Step 3: Divide the entire equation by 5 to simplify further: m^4 - 10m^3 + 33m^2 - 36m = 0

Step 4: Now, we can apply Vieta's formulas for a quartic equation, which state that for an equation of the form ax^4 + bx^3 + cx^2 + dx + e = 0, the sum of the roots is given by -b/a.

In our case, the coefficient of m^3 is -10, and the coefficient of m^4 is 1. So, the sum of the roots of the given equation is:

Sum of roots = -(-10)/1 = 10

Therefore, the sum of the roots of the equation 5(m-1)(m-3)(m-2)(m-4) - 120 = 0 is 10.