Forming a quadratic equation given roots that are fractions?

[Guest]

Could anyone explain how to form a quadratic equation given roots that are fractions ?
For example 1/2 , -2/3

[Guest]

Let $x_1=\frac{1}{2}$
$x_2=-\frac{2}{3}$

Quadratic equation is in the form of:
$ax^2+bx+c=0$

From Vieta's formulas we have:
$x_1+x_2=-\frac{b}{a}$
$x_1x_2=\frac{c}{a}$

Let we form an equation where $a=1$
And $x_1+x_2=-b \Rightarrow \frac{1}{2}+(-\frac{2}{3}) = -b$
$ \frac{1}{2}-\frac{2}{3} = \frac{3}{6}-\frac{4}{6} = -\frac{1}{6}$
$-b= -\frac{1}{6}$
$b=\frac{1}{6}$

Then
$x_1x_2 = c$
$\frac{1}{2}\cdot(-\frac{2}{3}) = -\frac{1}{3}$

So the equation is
$x^2+\frac{1}{6}x-\frac{1}{3}=0$
or $6x^2+x-2=0$

[Guest]

Equivalently, a and b are roots of the quadratic equation (x- a)(x- b)= 0. With a= 1/2 and b= -2/3 that is (x- 1/2)(x+ 2/3)= x^2- (1/2)x+ (2/3)x- 1/3= x^2+ (4/6- 3/6)x- 1/3= x^2+ (1/6)x- 1/3= 0.

That is a perfectly valid quadratic equation but if you really don't like fractions, multiply by 6 to get
6x^2+ x- 2= 0.

[Guest]

Equivalently, a and b are roots of the quadratic equation (x- a)(x- b)= 0. With a= 1/2 and b= -2/3 that is (x- 1/2)(x+ 2/3)= x^2- (1/2)x+ (2/3)x- 1/3= x^2+ (4/6- 3/6)x- 1/3= x^2+ (1/6)x- 1/3= 0.

That is a perfectly valid quadratic equation but if you really don't like fractions, multiply by 6 to get
6x^2+ x- 2= 0.

Good!