Answer:
The product of its roots is 6.
Step-by-step explanation:
Let [tex]m\cdot x^{2}-(6\cdot m +1)\cdot x+(m^{3}+m^{2}+4) = 0[/tex], all roots are calculated by Quadratic Formula:
[tex]r_{1} = \frac{6\cdot m +1 + \sqrt{(6\cdot m + 1)^{2}-4\cdot m\cdot (m^{3}+m^{2}+4)}}{2\cdot m}[/tex] (1)
[tex]r_{2} = \frac{6\cdot m +1 - \sqrt{(6\cdot m + 1)^{2}-4\cdot m\cdot (m^{3}+m^{2}+4)}}{2\cdot m}[/tex] (2)
According to statement, we know that:
[tex]r_{1}+r_{2} = 7[/tex] (3)
By applying (1) and (2) in (3), we have the following expression:
[tex]\frac{6\cdot m +1}{m} = 7[/tex]
[tex]6\cdot m + 1 = 7\cdot m[/tex]
[tex]m = 1[/tex]
If we know that [tex]m = 1[/tex], then the roots of the polynomial are, respectively:
[tex]x^{2}-7\cdot x +6 = 0[/tex]
[tex]r_{1} = 6[/tex]
[tex]r_{2} = 1[/tex]
And the product of the roots is:
[tex]r_{1}\cdot r_{2} = 6[/tex]
The product of its roots is 6.
