Quadratic equation having roots $a, b$ such that $a+b = 2m_1$ and $ab=4m_1^2 -3m_2$

Question

Suppose that $a+b = 2m_{1}$ and $ab = 4m_{1}^{2}-3m_{2}$. Why is the quadratic equation $$y^{2}-2m_{1}y+(4m_{1}^{2}-3m_{2})=0$$ instead of $$y^{2}+2m_{1}y+(4m_{1}^{2}-3m_{2})=0$$

In other words, why is it $-2m_{1}y$ instead of $2m_{1}y$ in the second term?

Answer 1

Because, if a quadratic equation has two roots $x_1$ and $x_2$, then one has $$ x^2+bx+c=(x-x_1)(x-x_2). $$ Comparing the coefficients at the equal powers yields $$ x_1+x_2=-b,\quad x_1 x_2=c, $$ note the minus sign.

Answer 2

if you have equation like this $x^2+b*x+c$ then it is called standard quadratic equation and by Vieta's Theorem,if $x_1$ and $x_2$ are roots of this equation then

$x_1+x_2=-b$

and

$x_1*x_2=c$

EDITED: so see it by example we know that if $x_1$,$x_2$ are roots then $(x^2+b*x+c)$=(x-$x_1$)*(x-$x_2$) then let us make operations on the right side,we have $x^2-x*(x_1+x_2)+x_1*x_2$

now if we look at $x^2+b*x+c$

we can see that $x_1+x_2=-b$ (to make - sign positive) and $x_1*x_2=c$

Answer 3

$a+b = x$ and $ab = y$

$$(a+b)^2 = x^2$$

$$a^2+2ab+b^2 = x^2$$

$$a^2+2ab+(x-a)^2 = x^2$$

$$a^2+2y+x^2-2xa+a^2 =x^2$$

$$2a^2+2y-2xa =x^2-x^2=0$$

$$2(a^2-xa+y) =0$$

Equation $(1)$ $$a^2-xa+y =0$$

if you put $a=x-b$ then

$$(x-b)^2-x(x-b)+y =0$$

$$x^2-2xb+b^2-x^2+xb+y =0$$

Equation $(2)$

$$b^2-xb+y =0$$

As you see Equation $(1)$ and Equation $(2)$ are the same Quadratic equation . $a$ and $b$ are roots of the Quadratic equation. As you see $x$ must be minus $y$ must be plus while writing the quadratic equation.