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How can I solve the following problem?

$4\cdot 2\sqrt{x}=x^2$

Andreas Caranti
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Hi there
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  • You can get the square root sign with \sqrt{x} ($\sqrt{x}$). If you want other roots, you can get them, too. For instance, the $n$-th root of $x$ is \sqrt[n]{x} ($\sqrt[n]{x}$). – Brian M. Scott Oct 29 '16 at 15:21
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    Wait, you meant $4 \sqrt{x} = x^{2}$? The current edit might be incorrect. – Andreas Caranti Oct 29 '16 at 15:22

3 Answers3

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Have you tried squaring both sides?

Aggelos Bessis
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  • Actually there are two possible solutions. Do you know what the other one is? Hint: square both sides to get $64x = x^4$. What's the next step? How do you know you can do the next step? – fleablood Oct 29 '16 at 15:42
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HINT: $4\cdot2=8=2^3$, and $\dfrac{x^2}{\sqrt{x}}=\dfrac{x^2}{x^{1/2}}=x^{2-\frac12}=\ldots$

Brian M. Scott
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    Don't forget $x = 0$! – fleablood Oct 29 '16 at 15:24
  • Oh.... "hint".... right. Not a solution. A hint to get to the solution.... – fleablood Oct 29 '16 at 15:25
  • @fleablood: I debated mentioning $x=0$, but I guess that now I don’t need to! – Brian M. Scott Oct 29 '16 at 15:27
  • I have solved it now, X = 4. But not on your way. I read from the above comment that I should square both sides.

    But on your way I can't seem to figure out how to continue from : 2^3 = x^1,5 I was trying with Log(8) / Log (1,5) but that wasn't correct either.

     – Hi there Oct 29 '16 at 15:34
  • @Hithere: The other method works just fine and is probably easier. What I was suggesting is that for $x\ne 0$ you get $2^3=x^{3/2}$, as you did; then you can easily take cube roots on both sides to get $2=x^{1/2}$, and squaring gives you $x=4$. Since we divided by $\sqrt{x}$, we also have to check the possibility that $x=0$, and of course it does turn out to give us another solution. – Brian M. Scott Oct 29 '16 at 15:36
  • Ah I see, alright. Thanks a lot for your help/time! – Hi there Oct 29 '16 at 15:38
  • @Hithere: You’re welcome! – Brian M. Scott Oct 29 '16 at 15:38
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$4*2\sqrt{x}=x^2$; for $\sqrt{x}$ to be meaningful $x \ge 0$.

$8x^{\frac 12} =(x^{\frac 12})^4$; if $x^{\frac 12}=0$ then $8*0=0^4$ and $x =0$

$8 = (x^{\frac 12})^3$; if $x^{\frac 12} \ne 0$.

$\sqrt[3]{8} = x^{\frac 12}$

$2 = x^{\frac 12}$

$2^2 = x$

$x=4$

So $x = 4$ or $x=0$.

fleablood
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