Question

Find the constant (a) such that ((x^2-3x+4)(2x^2+ax+7) = 2x^4-11x^3+30x^2-41x+28.

a. a = 5
b. a = -5
c. a = 4
d. a = -4

Answer 1

To find the constant (a), we need to expand and simplify the given quadratic equation ((x^2 - 3x + 4)(2x^2 + ax + 7)). By comparing the coefficients with the given equation 2x^4 - 11x^3 + 30x^2 - 41x + 28, we can determine the value of a to be 0.75.

Step-by-step explanation:

To find the constant (a), we need to expand the equation ((x^2 - 3x + 4)(2x^2 + ax + 7)) and simplify it to match the given equation 2x^4 - 11x^3 + 30x^2 - 41x + 28.

Multiplying the binomials gives us: 2x^4 - 11x^3 + (4a - 3)x^2 + (7a - 12)x + 28.

Comparing the coefficients of the two equations, we can equate the like terms. By comparing the constant terms, we can determine that 4a - 3 = 0, which implies a = 3/4 or a = 0.75.

Therefore, the correct option is a = 0.75.