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I am very terrible at some aspects of algebra and I would like to ask how to solve this problem (It's actually only a small part of a larger physics problem). I've looked up the laws of exponents and from what I can tell I cannot easily seperate the $x$ from the $0.6$. At this point I am quite confused and would appreciate a step by step approach:

Solve for $x$:

$$(x+0.6)^2 = 1.4x^2$$

Community
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PedroPovedaQ
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4 Answers4

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$$(x+0.6)^2 = 1.4x^2$$ $$x^2+1.2x+0.36=1.4x^2$$ $$-0.4x^2+1.2x+0.36=0$$

Now use the ABC-formula to solve for x.

Bob
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You can start by taking the square root of each side of the equation. $$\sqrt{(x + 0.6)^2} = \sqrt{1.4x^2} =(x+0.6) = \sqrt{1.4}x$$

Then solve for $x$, and evaluate:

$$(\sqrt{1.4} - 1)x = 0.6 \implies x = \dfrac{0.6}{\sqrt{1.4} - 1} \approx 3.275$$

Note that this method gives you only the positive solution for $x$ (there is also a negative solution), but I suspect, given the solution you already have, is the only one applicable given the physic's context of the problem.

The alternative (and mathematically more sound) approach is to expand the term on the left hand side, and solve for the resulting quadratic equation:

$$ \begin{align} (x+0.6)^2 &= 1.4x^2\\ \\ x^2+1.2x+0.36 &=1.4x^2\\ \\ -0.4x^2+1.2x+0.36&=0 \end{align} $$

Now, you can use the quadratic formula to obtain both the positive solution and the negative solution: $x \approx -0.275$ and $x\approx 3.275$.

amWhy
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  • But remember the square root may be negative on one or both sides – Ross Millikan Jan 31 '13 at 19:41
  • @Ross Yes, I commented on that following the method used. I assumed, given the solution he was working with, that the negative root/solution did not apply (physics context). Generally, in a strictly algebra-precal context, I'd have suggested solving the quadratic to find both real solutions, one of which is, indeed, negative. – amWhy Jan 31 '13 at 19:52
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$$(x+0,6)^{2}=1,4x^2$$ $$x^2+1,2x+0,36=1,4x^2$$ $$0,4x^2-1,2x-0,36=0$$ $$0,4(x^2-3x-9)=0$$ $$x^2-3x-9=0$$

then we get $x=\frac{3+3\sqrt{5}}{2}$ or $x=\frac{3-3\sqrt{5}}{2}$

i think it can be an answer without calculate the approximation value

A Ricko Maulidar
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$(x+0.6)^2 = 1.4x^2$ is not really a quadratic equation because $a^2=b^2$ iff $a=\pm b$, since $a^2-b^2=(a-b)(a+b)$.

Hence from $(x+0.6)^2 = 1.4x^2$ we get $x+0.6 = \pm \sqrt{1.4}\,x$ and so $x=0.6/(1\pm\sqrt{1.4})$.

lhf
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