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If the difference of p(x)=ax^4/5 + x^3/3 -2x^2/5 + bx/7 -1 and q(x)= x^4/5 + x^3/4 -1/3 x^2 +1/2 x + 3 is h(x) =1/5 x^4 + 1/12 x^3 - 1/15 x^2 + 5/14 x - C then solve the polynomial f(x)= ax^2 + bx + C=0​

Respuesta :

Answer:

x = -2, -1.

Step-by-step explanation:

p(x)=ax^4/5 + x^3/3 -2x^2/5 + bx/7 -1  

q(x)= x^4/5 + x^3/4 -1/3 x^2 +1/2 x + 3

Difference =  

(a-1)x^4/5 -  x^3/12 - 1/3x^2/ 5 + (b - 7/2) x / 7 - 4

h(x) =1/5 x^4 + 1/12x^3 - 1/15 x^2 + 5/14 x - C

Comparing coefficients:

(a - 1)/5 = 1/ 5

---> a = 2

(b - 7/2)/7 = 5/14

--> 2(b - 7/2) = 5

--> b - 7/2 = 5/2

--> b = 6

C = 4

So we have the equation

2x^2 + 6x + 4 = 0

--> x^2 + 3x + 2 = 0

--> (x + 1)(x + 2) = 0

---> x = -2, -1.