Solve for $k$ when the equation has equal roots












1












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UPDATE: Solved thanks to turkyhundt and jimbo



The mathematical question is as follows:




Calculate the value of $k$ for which $2x^2 + 4x - k = 0$ has equal roots.




My working solves it to equal $-2$, but if we then put this back into the formula this is, of course, impossible. I use the discriminate of the quadratic equation, $b^2-4ac$.






quadratics discriminant







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  • $begingroup$
    Plugging in $-2$ gives you $2x^2+4x+2=0$ or $x^2+2x+1=0$ which does indeed give two equal roots. So I think you are correct that $k=-2$. As long as we are clear that the equation has a $-k$
    $endgroup$
    – turkeyhundt
    Oct 31 '15 at 23:30










  • $begingroup$
    @turkeyhundt If you put this back into b^2-4ac it equals 32.
    $endgroup$
    – Wonderer
    Oct 31 '15 at 23:32










  • $begingroup$
    $c=-k$, $k=-2$, so $b^2-4ac=4^2-(4)(2)(-(-2))=4^2-(4)(2)(2)=0$
    $endgroup$
    – turkeyhundt
    Oct 31 '15 at 23:34












  • $begingroup$
    @turkeyhundt Thank you.
    $endgroup$
    – Wonderer
    Oct 31 '15 at 23:38
















1












$begingroup$


UPDATE: Solved thanks to turkyhundt and jimbo



The mathematical question is as follows:




Calculate the value of $k$ for which $2x^2 + 4x - k = 0$ has equal roots.




My working solves it to equal $-2$, but if we then put this back into the formula this is, of course, impossible. I use the discriminate of the quadratic equation, $b^2-4ac$.






quadratics discriminant







share|cite|improve this question











$endgroup$












  • $begingroup$
    Plugging in $-2$ gives you $2x^2+4x+2=0$ or $x^2+2x+1=0$ which does indeed give two equal roots. So I think you are correct that $k=-2$. As long as we are clear that the equation has a $-k$
    $endgroup$
    – turkeyhundt
    Oct 31 '15 at 23:30










  • $begingroup$
    @turkeyhundt If you put this back into b^2-4ac it equals 32.
    $endgroup$
    – Wonderer
    Oct 31 '15 at 23:32










  • $begingroup$
    $c=-k$, $k=-2$, so $b^2-4ac=4^2-(4)(2)(-(-2))=4^2-(4)(2)(2)=0$
    $endgroup$
    – turkeyhundt
    Oct 31 '15 at 23:34












  • $begingroup$
    @turkeyhundt Thank you.
    $endgroup$
    – Wonderer
    Oct 31 '15 at 23:38














1












1








1





$begingroup$


UPDATE: Solved thanks to turkyhundt and jimbo



The mathematical question is as follows:




Calculate the value of $k$ for which $2x^2 + 4x - k = 0$ has equal roots.




My working solves it to equal $-2$, but if we then put this back into the formula this is, of course, impossible. I use the discriminate of the quadratic equation, $b^2-4ac$.






quadratics discriminant







share|cite|improve this question











$endgroup$




UPDATE: Solved thanks to turkyhundt and jimbo



The mathematical question is as follows:




Calculate the value of $k$ for which $2x^2 + 4x - k = 0$ has equal roots.




My working solves it to equal $-2$, but if we then put this back into the formula this is, of course, impossible. I use the discriminate of the quadratic equation, $b^2-4ac$.






quadratics discriminant




quadratics discriminant






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 9:00









Pablo

34813




34813










asked Oct 31 '15 at 23:26









WondererWonderer

10116




10116












  • $begingroup$
    Plugging in $-2$ gives you $2x^2+4x+2=0$ or $x^2+2x+1=0$ which does indeed give two equal roots. So I think you are correct that $k=-2$. As long as we are clear that the equation has a $-k$
    $endgroup$
    – turkeyhundt
    Oct 31 '15 at 23:30










  • $begingroup$
    @turkeyhundt If you put this back into b^2-4ac it equals 32.
    $endgroup$
    – Wonderer
    Oct 31 '15 at 23:32










  • $begingroup$
    $c=-k$, $k=-2$, so $b^2-4ac=4^2-(4)(2)(-(-2))=4^2-(4)(2)(2)=0$
    $endgroup$
    – turkeyhundt
    Oct 31 '15 at 23:34












  • $begingroup$
    @turkeyhundt Thank you.
    $endgroup$
    – Wonderer
    Oct 31 '15 at 23:38


















  • $begingroup$
    Plugging in $-2$ gives you $2x^2+4x+2=0$ or $x^2+2x+1=0$ which does indeed give two equal roots. So I think you are correct that $k=-2$. As long as we are clear that the equation has a $-k$
    $endgroup$
    – turkeyhundt
    Oct 31 '15 at 23:30










  • $begingroup$
    @turkeyhundt If you put this back into b^2-4ac it equals 32.
    $endgroup$
    – Wonderer
    Oct 31 '15 at 23:32










  • $begingroup$
    $c=-k$, $k=-2$, so $b^2-4ac=4^2-(4)(2)(-(-2))=4^2-(4)(2)(2)=0$
    $endgroup$
    – turkeyhundt
    Oct 31 '15 at 23:34












  • $begingroup$
    @turkeyhundt Thank you.
    $endgroup$
    – Wonderer
    Oct 31 '15 at 23:38
















$begingroup$
Plugging in $-2$ gives you $2x^2+4x+2=0$ or $x^2+2x+1=0$ which does indeed give two equal roots. So I think you are correct that $k=-2$. As long as we are clear that the equation has a $-k$
$endgroup$
– turkeyhundt
Oct 31 '15 at 23:30




$begingroup$
Plugging in $-2$ gives you $2x^2+4x+2=0$ or $x^2+2x+1=0$ which does indeed give two equal roots. So I think you are correct that $k=-2$. As long as we are clear that the equation has a $-k$
$endgroup$
– turkeyhundt
Oct 31 '15 at 23:30












$begingroup$
@turkeyhundt If you put this back into b^2-4ac it equals 32.
$endgroup$
– Wonderer
Oct 31 '15 at 23:32




$begingroup$
@turkeyhundt If you put this back into b^2-4ac it equals 32.
$endgroup$
– Wonderer
Oct 31 '15 at 23:32












$begingroup$
$c=-k$, $k=-2$, so $b^2-4ac=4^2-(4)(2)(-(-2))=4^2-(4)(2)(2)=0$
$endgroup$
– turkeyhundt
Oct 31 '15 at 23:34






$begingroup$
$c=-k$, $k=-2$, so $b^2-4ac=4^2-(4)(2)(-(-2))=4^2-(4)(2)(2)=0$
$endgroup$
– turkeyhundt
Oct 31 '15 at 23:34














$begingroup$
@turkeyhundt Thank you.
$endgroup$
– Wonderer
Oct 31 '15 at 23:38




$begingroup$
@turkeyhundt Thank you.
$endgroup$
– Wonderer
Oct 31 '15 at 23:38










2 Answers
2






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$begingroup$

$b^2-4ac=4^2-4(2)(-k)=16+8k=0$ imply $k=-2$






share|cite|improve this answer









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    0












    $begingroup$

    $textbf{Another approach:}$
    begin{align}
    2x^2 + 4x - k &= 0\
    x^2 +4x-frac{k}{2}&= 0 \
    (x+1)^2-1 &=frac{k}{2} \
    0 &=frac{k}{2}+1 \
    k &=-2
    end{align}






    share|cite|improve this answer









    $endgroup$














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      2 Answers
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      2 Answers
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      2












      $begingroup$

      $b^2-4ac=4^2-4(2)(-k)=16+8k=0$ imply $k=-2$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $b^2-4ac=4^2-4(2)(-k)=16+8k=0$ imply $k=-2$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $b^2-4ac=4^2-4(2)(-k)=16+8k=0$ imply $k=-2$






          share|cite|improve this answer









          $endgroup$



          $b^2-4ac=4^2-4(2)(-k)=16+8k=0$ imply $k=-2$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 31 '15 at 23:28









          jimbojimbo

          1,635713




          1,635713























              0












              $begingroup$

              $textbf{Another approach:}$
              begin{align}
              2x^2 + 4x - k &= 0\
              x^2 +4x-frac{k}{2}&= 0 \
              (x+1)^2-1 &=frac{k}{2} \
              0 &=frac{k}{2}+1 \
              k &=-2
              end{align}






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $textbf{Another approach:}$
                begin{align}
                2x^2 + 4x - k &= 0\
                x^2 +4x-frac{k}{2}&= 0 \
                (x+1)^2-1 &=frac{k}{2} \
                0 &=frac{k}{2}+1 \
                k &=-2
                end{align}






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $textbf{Another approach:}$
                  begin{align}
                  2x^2 + 4x - k &= 0\
                  x^2 +4x-frac{k}{2}&= 0 \
                  (x+1)^2-1 &=frac{k}{2} \
                  0 &=frac{k}{2}+1 \
                  k &=-2
                  end{align}






                  share|cite|improve this answer









                  $endgroup$



                  $textbf{Another approach:}$
                  begin{align}
                  2x^2 + 4x - k &= 0\
                  x^2 +4x-frac{k}{2}&= 0 \
                  (x+1)^2-1 &=frac{k}{2} \
                  0 &=frac{k}{2}+1 \
                  k &=-2
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 9:51









                  1Spectre11Spectre1

                  1069




                  1069






























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