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# Help!

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The difference of the roots of the quadratic equation $x^2 + bx + c = 0$ is $|b - 2c|$. If $c \neq 0$, then find $c$ in terms of $b$.

Nov 17, 2020

#1
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Hello Guest!

By the quadratic formula, $x = \frac{-b + \sqrt{b^2 - 4c}}{2}, \frac{-b - \sqrt{b^2 - 4c}}{2}$. The difference of these is $\frac{2\sqrt{b^2 - 4c}}{2} = \sqrt{b^2 - 4c}$. Setting this equal to $|b - 2c|$, it follows that (after squaring) $b^2 - 4c = (b-2c)^2 = b^2 + 4c^2 - 4bc$. Thus$$0 = 4c^2 + 4c - 4bc = 4c(c - b + 1).$$As $c \neq 0$, it follows that $c = \boxed{b - 1}$.

Hope this helped!

Nov 17, 2020

#1
+97
+2

Hello Guest!

By the quadratic formula, $x = \frac{-b + \sqrt{b^2 - 4c}}{2}, \frac{-b - \sqrt{b^2 - 4c}}{2}$. The difference of these is $\frac{2\sqrt{b^2 - 4c}}{2} = \sqrt{b^2 - 4c}$. Setting this equal to $|b - 2c|$, it follows that (after squaring) $b^2 - 4c = (b-2c)^2 = b^2 + 4c^2 - 4bc$. Thus$$0 = 4c^2 + 4c - 4bc = 4c(c - b + 1).$$As $c \neq 0$, it follows that $c = \boxed{b - 1}$.

Hope this helped!

ETERNITY Nov 17, 2020
#2
+114221
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Very nice, Eternity    !!!!

CPhill  Nov 17, 2020