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I am to solve for x: $7x^2+3x=0$

I'm aware that there are multiple approaches to solving a quadratic. In this case, since there is no constant term I decided to go the completing the square route. I know from my textbooks answer that the solutions are $x=0$ and $x=-\frac{3}{7}$.

Here is how far I got:

$7x^2+3x=0$ # want to have leading coefficient 1 not 7

$x^2 + \frac{3}{7}x=0$

take 1/2 of the linear coefficient and then square it:

$\frac{1}{2}*\frac{3}{7}=\frac{3}{14}$

Then square it:

$(\frac{3}{14})^2$ = $\frac{9}{196}$

Add this term to both sides of my equation:

$x^2 + \frac{3}{7}x + \frac{9}{196}=\frac{9}{196}$

This is where I get stuck. Apparently I should be able to factor as a perfect square the left hand side of the equation. Perhaps because I'm working with fractions I cannot see how to do that? How can I turn $x^2 + \frac{3}{7}x + \frac{9}{196}$ into the form $(x+n)^2$?

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    $\begingroup$ You did not square the term. The square of the term should be $\dfrac{9}{14^2}$ so that we get $\left( x + \dfrac{3}{14} \right)^2 = \dfrac{9}{14^2}$. $\endgroup$ – Aniruddha Deshmukh May 1 at 5:46
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    $\begingroup$ You have forgot squaring $\frac{3}{14}$. You should add $\left(\frac{3}{14}\right)^2$ on both sides. $\endgroup$ – Yuta May 1 at 5:48
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    $\begingroup$ Doesn't seem like you squared the $3/14$. $\endgroup$ – coffeemath May 1 at 5:48
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    $\begingroup$ since there is no constant term I decided to go the completing the square route --- When there is no constant term, the route you should be taking is by factoring, which is super-easy when there is no constant term. $\endgroup$ – Dave L. Renfro May 1 at 5:51
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    $\begingroup$ Take out $x$ as common factor, what do you get then? Try expanding $(x+3/14)^2$ $\endgroup$ – rhombicosicodecahedron May 1 at 6:01
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How can I turn $x^2 + \frac{3}{7}x + \frac{9}{196}$ into the form $(x+n)^2$?

A hint: $n$ is always half of the linear coefficient. You've already calculated that this is $3/14$.

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$$7x^2+3x=0 \iff 7(x^2+(3/7)x)=0\iff$$ $$\iff 7 (\;(x+(1/2)(3/7)\,)^2 -(\,(1/2)(3/7)\,)^2\;)=0\iff $$ $$\iff (x+(1/2)(3/7)\,)^2-(\,(1/2)(3/7)\,)^2=0 \iff$$ $$\iff (x+(1/2)(3/7)\,)^2=(\,(1/2)(3/7)\,)^2\iff$$ $$\iff x+(1/2)(3/7)=\pm (1/2)(3/7)\iff$$ $$\iff (\;(x+(1/2)(3/7)=(1/2)(3/7)\; \lor \;x+(1/2)(3/7)=-(1/2)(3/7)\;) \iff$$ $$\iff (\,x=0\,\lor \,x=-3/7\,).$$

Of course if $A\ne 0$ then $Ax^2+Bx=0 \iff A(x)(x+B/A)=0 \iff (x=0 \lor x+B/A=0)\iff (x=0\lor x=-B/A).$

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