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$x^2 + \sqrt{2}x = \frac{1}{2}$

I need to find the real solutions for this equation and write it in this form:

$$\frac{-\sqrt{A} \pm B}{C}$$

So when I work the problem out with the quadratic equation I get: $x = 0.118121$

I had no idea how to even put that in the form described so I assumed they didn't want me to solve but just put it in that form so I put it as follows:

$A = x^2$

$B = \sqrt{2}x$

$C = \frac{1}{2}$

That was wrong also. I'm not really sure what they're asking me to do. I know the answer is: $x = 0.118121$ but is it even possible to put it in the form described? Any help would be appreciated.

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    $\begingroup$ Put the calculator aside and use the quadratic formula. $\endgroup$ Jun 13, 2016 at 6:32
  • $\begingroup$ I did use the quadratic formula and got that answer? They're asking for solutions but typically those solutions come in 2 sets of "x". for example x=2, x=4. but it never comes out as A = 1, B=5,C=9. or anything like that. And again as I described, I re-arranged it in the standard form and then plugged in the A,B,and C to satisfying the form they're asking but it was still wrong. $\endgroup$ Jun 13, 2016 at 6:57
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    $\begingroup$ Do they accept the (correct) answer by Angelo Mark? $\endgroup$ Jun 13, 2016 at 13:23
  • $\begingroup$ @AndréNicolas Yes. I thought they were asking for the "x" values(because that's the solution to the equation) and then have that converted int o the form above, but it never occurred to me that they wanted a condensed version of it and that it would eventually evolve into the form that they wanted. $\endgroup$ Jun 13, 2016 at 20:41

2 Answers 2

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Form of any quadratic Equation is $ax^2+bx+c=0$

Here for you $ a=1 , b= \sqrt{2}$ and $c = -\frac{1}{2}$, $$x^2 + \sqrt{2}x + \left(-\frac{1}{2} \right)=0$$

Solution is $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$x=\frac{-\sqrt{2}\pm \sqrt{(\sqrt{2})^2-4\cdot 1 \cdot \left(-\frac{1}{2}\right)}}{2 \cdot 1}$$

$$x=\frac{-\sqrt{2}\pm \sqrt{2+2}}{2 }$$

$$x=\frac{-\sqrt{2}\pm \sqrt{4}}{2 }$$

$$x=\frac{-\sqrt{2}\pm 2}{2 }$$

So $$A=2 ,B =2 ,C=2$$

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  • $\begingroup$ Ok I did and got x=0.2928932188134524,−1.7071067811865475 I'm still not sure how to put that in the form they want? $\endgroup$ Jun 13, 2016 at 6:47
  • $\begingroup$ @user3882522 Now check the answer :) $\endgroup$ Jun 13, 2016 at 7:03
  • $\begingroup$ Wow i feel stupid lol. Thanks bro. $\endgroup$ Jun 13, 2016 at 7:34
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Hint: $x^2+\sqrt{2}x=\frac{1}{2}\implies2x^2+2\sqrt{2}x-1=0$ Now use the general way to solve an quadratic.

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