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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.

Select all that apply If x2baxca0 then The sum of its roots ba The difference of its roots ba The product of its roots caThe division of its roots ca I can sele class=

Respuesta :

unicorn3125

Answer:

             The first and the third

Step-by-step explanation:

[tex]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}[/tex]

[tex]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}=[/tex]

[tex]=\frac{b^2+\sqrt{b^2-4ac}-2ac}{2ac}[/tex]

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