Finding the value of a constant given an equation where the sum of the roots is -3

Question

I am to find the value of h given the equation 3hx^2 - 2x +5xh = 3.
The sum of the roots of the polynomial is -3.
I am having trouble finding the roots of the equation given the hindrance of h. How do I find the roots such that I may find the value of h?

Answer 1

$3hx^2-2x+5hx=3\implies$

$3hx^2+(5h-2)x-3=0\implies$

$x_{1,2}=\frac{2-5h\pm\sqrt{(5h-2)^2+4\cdot3h\cdot3}}{2\cdot3h}\implies$

$x_{1,2}=\frac{2-5h\pm\sqrt{25h^2+16h+4}}{6h}$

---

$x_1+x_2=-3\implies$

$\frac{2-5h+\sqrt{25h^2+16h+4}}{6h}+\frac{2-5h-\sqrt{25h^2+16h+4}}{6h}=-3\implies$

$\frac{2-5h+2-5h}{6h}=-3\implies$

$4-10h=-18h\implies$

$8h=-4\implies$

$h=-\frac{1}{2}\implies$

Answer 2

First rewrite your equation in standard form as follows:$$3hx^2-2x+5xh=3$$$$\therefore3hx^2+(5h-2)x-3=0$$Assuming $h\ne0$ we can divide it all by $3h$ to get:$$x^2+\frac{(5h-2)}{3h}x-\frac{1}{h}=0\tag{1}$$Now, if a quadratic equation has roots $r_1$ and $r_2$, then it can be written as:$$(x-r_1)(x-r_2)=0$$$$\therefore x^2-(r_1+r_2)x+r_1r_2=0\tag{2}$$You are told that the sum of the roots is $-3$. Therefore you know that:$$r_1+r_2=-3\tag{3}$$You should now be able to use equations (1), (2) and (3) to get your answer.