Question

Select all that apply. If x^2+b/ax+c/a=0 ; then: The sum of its roots = -b/a? The difference of its roots =-b/a? The product of its roots = c/a?The division of its roots = c/a? I can select multiple.

Answer 1

Answer:

             The first and the third

Step-by-step explanation:

$x^2+\frac bax+\frac ca=0\\\\ ax^2+bx+c=0\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\qquad\quad x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\\\x_1+x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}+\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{-2b}{2a}=\dfrac{-b}a\\\\\\x_1\cdot x_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\cdot\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\\\\{}\ \ =\dfrac{b^2-b\sqrt{b^2-4ac}+b\sqrt{b^2-4ac}-(\sqrt{b^2-4ac})^2}{2a}=\dfrac{b^2-(b^2-4ac)}{4a^2}=\\\\{}\ \ =\dfrac{b^2-b^2+4ac}{4a^2}=\dfrac{4ac}{4a^2}=\dfrac{c}{a}$

$x_1-x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}-\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-2\sqrt{b^2-4ac}}{2a}=\frac{-\sqrt{b^2-4ac}}{a}\\\\\\x_1\div x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\div\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{-b-\sqrt{b^2-4ac}}{2a}\,\cdot\,\frac{2a}{-b+\sqrt{b^2-4ac}}=\\\\=\frac{-b-\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}}=\frac{b+\sqrt{b^2-4ac}}{b-\sqrt{b^2-4ac}}=\frac{b^2+2\sqrt{b^2-4ac}+b^2-4ac}{b^2-b^2+4ac}=\frac{2b^2+2\sqrt{b^2-4ac}-4ac}{4ac}=$

$=\frac{b^2+\sqrt{b^2-4ac}-2ac}{2ac}$