Q:

the difference between roots of the quadratic equation x^2+x+c=0 is 6. find c.

Accepted Solution

A:
Answer:[tex]\displaystyle c = -\frac{35}{4} = -8.75[/tex].Step-by-step explanation:Let the smaller root to this equation be [tex]m[/tex]. The larger one will equal [tex]m + 6[/tex].By the factor theorem, this equation is equivalent to [tex]a(x - m)(x - (m+6))= 0[/tex], where [tex]a \ne 0[/tex].Expand this expression:[tex]a\cdot x^{2} - a(2m + 6)\cdot x + a(m^{2} + 6m) =0[/tex].This equation and the one in the question shall differ only by the multiple of a non-zero constant. It will be helpful if that constant is equal to [tex]1[/tex]. That way, all constants in the two equations will be equal; [tex](m^{2} + 6m)[/tex] will Β be equal to [tex]c[/tex].Compare this equation and the one in the question:The coefficient of [tex]x^{2}[/tex] in the question is [tex]1[/tex] (which is omitted.) The coefficient of [tex]x^{2}[/tex] in this equation is [tex]a[/tex]. If all corresponding coefficients in the two equations are equal to each other, these two coefficients shall also be equal to each other. Therefore [tex]a = 1[/tex].This equation will become:[tex]x^{2} - (2m + 6)\cdot x + (m^{2} + 6m) =0[/tex].Similarly, for the coefficient of [tex]x[/tex], [tex]\displaystyle -(2m +6) = 1[/tex].[tex]\displaystyle m = -\frac{7}{2}[/tex].This equation will become: [tex]x^{2} + x + \underbrace{\left(-\frac{35}{4}\right)}_{c} =0[/tex].[tex]c[/tex] is the value of the constant term of this quadratic equation.