Question

Find the quadratic equation whose roots are -2/3 and -1/4

Answer 1

Let the quadratic equation be

$$ax^2+bx+c=0, a\not=0$$

By the Vieta's theorem

$$x_1+x_2=-b/a$$

$$x_1x_2=c/a$$

Let $a=1.$ Then

$$b=-(-2/3-1/4)=11/12$$

$$c=(-2/3)(-1/4)=1/6$$

The quadratic equation is

$$x^2+\dfrac{11}{12}x+\dfrac{1}{6}=0$$

Or

$$12x^2+11x+2=0$$