sonokmahapatra sonokmahapatra
  • 19-09-2016
  • Mathematics
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The quadratic equation $y = x^2 + 5x + c$ has solutions in the form of $x = \frac{-5 \pm \sqrt{c}}{2}$. What is the value of $c$?

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dalendrk
dalendrk dalendrk
  • 19-09-2016
[tex]ax^2+bx+c=0\\\\x_1=\dfrac{-b-\sqrt{\Delta}}{2a}\ and\ x_2=\dfrac{-b+\sqrt{\Delta}}{2a}\\\\\Delta=b^2-4ac[/tex]

[tex]y=x^2+5x+c\\\\a=1;\ b=5;\ c=c\\\\x=\dfrac{-5\pm\sqrt\Delta}{2\cdot1}\\\\\Delta=5^2-4\cdot1\cdot c=25-4c\\\\Therefore:x=\dfrac{-5\pm\sqrt{25-4c}}{2}[/tex]
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