# Quadratic formula problem $(x+2)^2-(2x+3)^2=0$

Solve $(x+2)^2-(2x+3)^2=0$ using the quadratic formula.

I have tried expanding the brackets in both and then simplifying but I get an equation that I can't then factorise.

Can I have a method please?

• Welcome to the site. What have you tried? Oct 13, 2014 at 19:19
• You can compute the squares to get an ordinary quadratic equation.
– mfl
Oct 13, 2014 at 19:19
• I have tried expanding the brackets in both and then simplifying but I get an equation that I can't then factorise Oct 13, 2014 at 19:30
• What equation have you obtained?
– mfl
Oct 13, 2014 at 19:31
• You have some problem with signs. Remember, $a-(b+c)=a-b-c.$ Check it and use the standard formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ to solve it.
– mfl
Oct 13, 2014 at 19:37

If you want to use the quadratic formula, at first you must solve the parentheses. The you would have $x^2+4x+4-4x^2-12x-9=0 \Leftrightarrow -3x^2-8x-5=0$.

$$x=\frac{8+\sqrt{8^2-4(3)(5)}}{-6}=-5/3$$

$$x=\frac{8-\sqrt{8^2-4(3)(5) })}{-6}=-1$$

You could also solve the problem imposing that $|x+2|=|2x+3|$, but except that you do it graphically, this process is too complicated.

Note that you do not have to use the quadratic formula, although it certainly is an option. I have no idea why Zero point says this method is "too complicated." Realize that the original equation is equivalent to $$(x+2)^2 = (2x+3)^2,$$ which is true if $\left|x+2\right|=\left|2x+3\right|$, and can be split up into two cases.

If $x + 2 = 2x + 3$, we get $x = -1$ and the statement is true.

If $x + 2 = - \left( 2x + 3 \right)$, we get $x = - \frac {5}{3}$, in which case the statement is also true.

So we have found our two solutions $x \in \left\{ -1, - \frac{5}{3} \right\}$. $\Box$