Answer:
The sum of the two solutions to the given equation is 1/3.
Step-by-step explanation:
Nature of Roots of a Second-Degree Equation
Given a quadratic equation:
[tex]ax^2 + bx + c = 0[/tex]
The roots of the equation are:
[tex]\displaystyle x_1=\frac{-b+ \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\displaystyle x_2=\frac{-b- \sqrt{b^2-4ac}}{2a}[/tex]
The sum of the roots is:
[tex]\displaystyle x_1+x_2=-\frac{b}{a}[/tex]
The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient.
We are given the equation:
[tex]3x^2 - x = 4[/tex]
Rearranging:
[tex]3x^2 - x - 4=0[/tex]
The coefficient b is b=-1, and the leading coefficient is a=3, thus the sum of the roots is:
[tex]\displaystyle x_1+x_2=-\frac{-1}{3}[/tex]
[tex]\displaystyle x_1+x_2=\frac{1}{3}[/tex]
The sum of the two solutions to the given equation is 1/3.
