Question

Writing a quadratic equation given the roots and the leading coefficient

roots 6, 4, and coefficient 5

Answer 1

`\bf x= \begin{cases} 6\\ 4 \end{cases}\implies \begin{cases} x=6\implies &x-6=0\\ x=4\implies &x-4=0 \end{cases} \\\\\\ (x-6)(x-4)=\stackrel{y}{0}\implies \stackrel{\mathbb{F~O~I~L}}{x^2-10x+24}=0\implies \stackrel{\textit{adding a common factor of 5}}{5(x^2-10x+24)=0} \\\\\\ 5x^2-50x+120=0\implies 5x^2-50x+120=y`

now, the common factor of 5 simply makes the parabola steeper, but the roots are the same, whilst the vertex of it changes