Question

Please helpppp me ! If the sum of the roots of the equation mx^2 - (6m+1)x+m^3+m^2+4=0 is equal to 7 , what is the product of its roots ? ( will give brainliest if answer correctly ) . ♡​

Answer 1

Answer:

The product of its roots is 6.

Step-by-step explanation:

Let $m\cdot x^{2}-(6\cdot m +1)\cdot x+(m^{3}+m^{2}+4) = 0$, all roots are calculated by Quadratic Formula:

$r_{1} = \frac{6\cdot m +1 + \sqrt{(6\cdot m + 1)^{2}-4\cdot m\cdot (m^{3}+m^{2}+4)}}{2\cdot m}$ (1)

$r_{2} = \frac{6\cdot m +1 - \sqrt{(6\cdot m + 1)^{2}-4\cdot m\cdot (m^{3}+m^{2}+4)}}{2\cdot m}$ (2)

According to statement, we know that:

$r_{1}+r_{2} = 7$ (3)

By applying (1) and (2) in (3), we have the following expression:

$\frac{6\cdot m +1}{m} = 7$

$6\cdot m + 1 = 7\cdot m$

$m = 1$

If we know that $m = 1$, then the roots of the polynomial are, respectively:

$x^{2}-7\cdot x +6 = 0$

$r_{1} = 6$

$r_{2} = 1$

And the product of the roots is:

$r_{1}\cdot r_{2} = 6$

The product of its roots is 6.