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Problem: solve (find roots): $ 7x^2-3x=0$.

How to find $c$ in order to solve using the Quadratic Formula?

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  • $\begingroup$ Do you mean $ax^2$? Anyway, in this case, $c=0$ $\endgroup$ Jun 17, 2015 at 23:24

5 Answers 5

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HINT Here $c=0$, so it is easier to simply factor to find the roots: $$x(7x-3)=0$$

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In your case $c = 0$.

$$ 7 x^2 -3x = 7 x^2 -3x + 0 = a x^2 + bx + c $$

So

$$ a = 7\quad b=-3\quad c=0 $$

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I think you mean $ax^{2}+bx+c=0$, here you simply have $c=0$. Further, there's no need for the quadratic formula, $x(7x-3)=0$

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$c$ is $0$ in this case. You can use the Quadratic formula right away to find $x=\frac{3\pm{\sqrt{9-0}}}{14}$ or factorize to get $f(x) = x(7x-3)$.

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Well, you don't need the quadratic formula here. Note that $7x^2-3x = x(7x-3)$, so this is equal to $0$ only when either $x = 0$ or $7x-3 = 0$, that is, when $x=0$ or $x = \frac{3}{7}$.

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