Solve $7x^2+3x=0$ by completing the square

Question

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 $(x+n)^2$?

Answer 1

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 $3/14$.