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"Find out the value of x by this equation: $(x+a)^2 = (2a-3x)^2$". (The answer should by the way, according to my book, be

$x1 = 0.25a$

$x2 = 1.5a$

Here's how far I've gotten:

$(x+a)^2 = (2a-3x)^2$

$x^2 + 2ax + a^2 = 4a^2 - 12ax + 9x^2$

$x^2 = 3a^2 + 9x^2 - 14ax$

After that I'm stuck. I haven't done any problem like this (ever) before, so I don't know what to do. I did try to "formulate" the equation differently to see if I would have an easier time solving it like this:

$-8x^2 + 14ax - 3a^2 = 0$

But I honestly don't know what do after that. I'd love it if someone could thoroughly explain the steps I need to do in order to get the value of $x$.

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    $\begingroup$ $a^2=b^2 $then $$a=\pm b$$ $\endgroup$
    – Khosrotash
    Jul 14, 2015 at 18:57
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    $\begingroup$ $ax^2+bx+c=0$ then $$x_1,x_2=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ $\endgroup$
    – Khosrotash
    Jul 14, 2015 at 19:00

3 Answers 3

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No quadratic equation really required here:

\begin{align*} (x + a)^2 = (2a - 3x)^2 & \iff x + a = \pm (2a - 3x)\\ & \iff x + a = \begin{cases} 2a - 3x\\ 3x - 2a \end{cases}\\ & \iff x = \begin{cases} \dfrac{a}{4}\\[0.2 mm] \dfrac{3a}{2} \end{cases} \end{align*}

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  • $\begingroup$ no quadratic equation really required ever $\endgroup$
    – pancini
    Jul 14, 2015 at 19:45
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    $\begingroup$ Certainly, but sometimes it's hidden and the formulae are handy. But here it's the basic case, which the whole theory relies upon. $\endgroup$
    – Bernard
    Jul 14, 2015 at 20:02
  • $\begingroup$ Right I guess I'm overcomplicating it $\endgroup$
    – pancini
    Jul 14, 2015 at 20:03
  • $\begingroup$ It's often the case for each of us. It should be a reflex to check if any solution may be simplified $\endgroup$
    – Bernard
    Jul 14, 2015 at 20:05
  • $\begingroup$ I used the align environment to align your equations since they were overlapping. This is certainly the simplest solution. $\endgroup$ Jul 15, 2015 at 9:32
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$8x^2-14ax+3a^2=0$

Use the quadratic equation

$x=\frac{14a±\sqrt{(14a)^2-4(8)(3a^2)}}{16}$

$=\frac{14a±10a}{16}$

etc.

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Treat symbol $a$ as if it is a constant.

$$ -8x^2 + 14ax - 3a^2 = ( 4 x -a)( -2 x + 3 a)= 0, $$

$$ x = ( 3 a/2, a/4 ) $$

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