Answer:
Step-by-step explanation:
Hello, I believe that we can consider a different from 0.
By definition of the roots we can write.
[tex]ax^2+bx+c=a(x-\alpha)(x-\beta)=a(x^2-(\alpha + \beta)x+\alpha \cdot \beta)\\\\\text{So we can say that:}\\\\\alpha + \beta = \dfrac{-b}{a}\\\\\alpha \cdot \beta=\dfrac{c}{a}\\\\\text{So the expected quadratic equation is}\\\\(x+\dfrac{b}{a})(x-\dfrac{c}{a})=0\\\\<=> \Large \boxed{\sf \bf \ (ax+b)(ax-c)=0 \ }[/tex]
Thank you
