#1**+3 **

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

**Express \(k\) as a common fraction.**

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

heureka Jun 10, 2019

#1**+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.**

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

heureka Jun 10, 2019