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$$2x^{ 2 }+8x+1=0$$

Move 1 to the other side of the equation:

$$2x^{ 2 }+8x\quad =-1$$

Divide both sides by 2 to get 1 as the leading coefficient:

$$x^{ 2 }+4x\quad =-\frac { 1 }{ 2 } $$

$$(\frac { 4 }{ 2 } )^{ 2 }$$

$$x^{ 2 }+4x+4\quad =-\frac { 1 }{ 2 } +4$$

$$(x+2)^{ 2 }=3.5$$

I get to this and it seems to be the wrong answer:

$$x={ -2 }\pm\sqrt { \frac { 7 }{ 2 } } $$

Claude Leibovici
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Cherry_Developer
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  • Looks good to me. Why do you think that's the wrong answer ? – David H Aug 23 '14 at 11:49
  • The textbook answer key saids $$x={ -2 }_{ - }^{ + }\sqrt { 14 } \ \quad \quad \quad $$ except radical 14 is divided by 2. Couldn't figure out how to put the /2 outside of the radical. – Cherry_Developer Aug 23 '14 at 11:53
  • Your answer key has a typo. – David H Aug 23 '14 at 11:56
  • You might be right because I just solved another problem using the quadratic formula and it has the negative signs mismatched. – Cherry_Developer Aug 23 '14 at 11:59
  • Your solution is good and the book is wrong ! This is not the first time. – Claude Leibovici Aug 23 '14 at 12:00
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    @Cherry_Developer I think you're saying the answer key says $\displaystyle -2 \pm \frac{\sqrt{14}}{2}$ in which case you're both right since $\displaystyle \sqrt\frac{7}{2} = \frac{\sqrt{14}}{2}.$ – Zoe H Aug 23 '14 at 12:01
  • That is what it saids. I just don't understand How one equals the other – Cherry_Developer Aug 23 '14 at 12:06
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    @Cherry_Developer $\displaystyle \sqrt{\frac{7}{2}} = \frac{\sqrt{7}}{\sqrt{2}} = \frac{\sqrt{7}}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{7\cdot2}}{\sqrt{2\cdot2}}\frac{\sqrt{14}}{2}$ – Zoe H Aug 23 '14 at 12:12

1 Answers1

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Your calculations are right..

$$(x+2)^2=\frac{7}{2} \Rightarrow x+2=\pm \sqrt{\frac{7}{2}} \Rightarrow x=-2 \pm \sqrt{\frac{7}{2}}=-2 \pm \frac{\sqrt{14}}{2}$$

So:

$$x=-2 \pm \frac{\sqrt{14}}{2}$$

evinda
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