What is the product of the roots of $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0" "$ where $a \neq 0$ $a != 0$ ?

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What is the product of the roots of $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0" "$ where $a \neq 0$ $a != 0$ ?

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Answer 1 · 2016-05-15T23:43:59

$\frac{c}{a}$ $c/a$

Explanation:

Method 1 - Direct multiplication

$\frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \cdot \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b-sqrt(b^2-4ac))/(2a) * (-b+sqrt(b^2-4ac))/(2a)$

$= \frac{{(- b)}^{2} - {(\sqrt{b^{2} - 4 a c})}^{2}}{4 a^{2}}$ $=((-b)^2-(sqrt(b^2-4ac))^2)/(4a^2)$

$= \frac{b^{2} - (b^{2} - 4 a c)}{4 a^{2}}$ $=(b^2-(b^2-4ac))/(4a^2)$

$= \frac{4 a c}{4 a^{2}}$ $=(4ac)/(4a^2)$

$= \frac{c}{a}$ $=c/a$

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Method 2 - Think about the roots

Call the two values of this expression $r_{1}$ $r_1$ and $r_{2}$ $r_2$ - the roots of the quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$ .

Then we have:

$a x^{2} + b x + c$ $ax^2+bx+c$

$= a (x - r_{1}) (x - r_{2})$ $= a(x-r_1)(x-r_2)$

$= a (x^{2} - (r_{1} + r_{2}) x + r_{1} r_{2})$ $= a(x^2-(r_1+r_2)x+r_1 r_2)$

$= a x^{2} - a (r_{1} + r_{2}) x + a r_{1} r_{2}$ $=ax^2-a(r_1 + r_2)x + a r_1 r_2$

Equating coefficients we find:

$c = a r_{1} r_{2}$ $c = a r_1 r_2$

So dividing both sides by $a$ $a$ we find:

$r_{1} r_{2} = \frac{c}{a}$ $r_1 r_2 = c/a$

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What is the product of the roots of $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0" "$ where $a \neq 0$ $a != 0$ ?

1 Answer

Explanation:

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What is the product of the roots of $a x^{2} + b x + c = 0$ $ax^2+bx+c = 0" "$ where $a \neq 0$ $a != 0$ ?