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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?

3 Answers 3

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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.

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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 $x_1*x_2=c$

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$a+b = x$ 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.