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deleted.

 Jun 10, 2019
edited by sinclairdragon428  Nov 20, 2019

Best Answer 

 #1
avatar+23862 
+3

Suppose that \(2x^2 - 5x + k = 0\) is a quadratic equation with one solution for \(x\).

Express \(k\) as a common fraction.

 

\(\begin{array}{|rcll|} \hline \mathbf{2x^2 - 5x + k} &=& \mathbf{0} \\\\ x &=& \dfrac{5\pm \sqrt{25-4\cdot 2k} }{2\cdot 2} \\ \hline 25-4\cdot 2k &=& 0 \\ 25-8k &=& 0 \\ 8k &=& 25 \\ \mathbf{k} &=& \mathbf{\dfrac{25}{8}} \\ \hline \end{array}\)

 

laugh

 Jun 10, 2019
 #1
avatar+23862 
+3
Best Answer

Suppose that \(2x^2 - 5x + k = 0\) is a quadratic equation with one solution for \(x\).

Express \(k\) as a common fraction.

 

\(\begin{array}{|rcll|} \hline \mathbf{2x^2 - 5x + k} &=& \mathbf{0} \\\\ x &=& \dfrac{5\pm \sqrt{25-4\cdot 2k} }{2\cdot 2} \\ \hline 25-4\cdot 2k &=& 0 \\ 25-8k &=& 0 \\ 8k &=& 25 \\ \mathbf{k} &=& \mathbf{\dfrac{25}{8}} \\ \hline \end{array}\)

 

laugh

heureka Jun 10, 2019

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