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?
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2$\begingroup$ Welcome to the site. What have you tried? $\endgroup$– N. F. TaussigOct 13, 2014 at 19:19
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1$\begingroup$ You can compute the squares to get an ordinary quadratic equation. $\endgroup$– mflOct 13, 2014 at 19:19
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1$\begingroup$ I have tried expanding the brackets in both and then simplifying but I get an equation that I can't then factorise $\endgroup$– CorinneOct 13, 2014 at 19:30
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$\begingroup$ What equation have you obtained? $\endgroup$– mflOct 13, 2014 at 19:31
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1$\begingroup$ 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. $\endgroup$– mflOct 13, 2014 at 19:37
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2 Answers
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If you want to use the quadratic formula, at first you must solve the parentheses. The you would have $x^2+4x+4-4x^2-12x-9=0 \Leftrightarrow -3x^2-8x-5=0$.
Using the quadratic formula(link) you get next results:
$$x=\frac{8+\sqrt{8^2-4(3)(5)}}{-6}=-5/3$$
$$x=\frac{8-\sqrt{8^2-4(3)(5) })}{-6}=-1$$
You could also solve the problem imposing that $|x+2|=|2x+3|$, but except that you do it graphically, this process is too complicated.
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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\} $. $\Box$
answered Oct 13, 2014 at 21:13