+0

0
190
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Find the sum of all integral values of $$c$$ with $$c\le 25$$ for which the equation $$y=x^2-7x-c$$ has two rational roots.

Jul 19, 2019

#1
+23872
+1

Find the sum of all integral values of $$c$$ with $$c\le 25$$ for which the equation $$y=x^2-7x-c$$ has two rational roots.

$$\begin{array}{|lrcll|} \hline & x^2-7x-c &=& 0 \\ & x &=& \dfrac{7\pm \sqrt{49-4(-c)}}{2} \\\\ & x &=& \dfrac{7\pm \sqrt{49+4c}}{2} \\\\ \text{Two rational roots} & 49+4c &>& 0 \\ & 4c &>& -49 \\ & c &>& -\frac{49}{4} \\ & \mathbf{ c } &>& \mathbf{-12.25} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \text{Integral values} & c &=& \{ -12,\ -11,\ -10,\ \ldots \ ,\ -1,\ 0,\ 1,\ \ldots \ ,\ 25\} \\\\ \text{The sum of all integral values of c with c\le 25} & &=& \left(\dfrac{-12+25}{2}\right)\cdot (25+13) \\ & &=& \dfrac{13\cdot 38}{2} \\ & &=& \mathbf{247} \\ \hline \end{array}$$

Jul 19, 2019
#2
0

That's wrong.  The right answer is $$-2.$$

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Jul 19, 2019
#3
+106535
+2

y = x^2  - 7x  - c

This will have two rational roots  when the discriminant  is a perfect square > 0

So

7^2  + 4c > 0

49  + 4c > 0

And   49 + 4c   will be a perfect square > 0  when   c  =  -6 , -10, -12, 0,  8, 18

So....the sum of the possible values of  c  =  -28 + 26   =   - 2

Jul 19, 2019