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Jun 10, 2019
edited by sinclairdragon428  Nov 20, 2019

#1
+23862
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Suppose that $$2x^2 - 5x + k = 0$$ is a quadratic equation with one solution for $$x$$.

Express $$k$$ as a common fraction.

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

Jun 10, 2019

#1
+23862
+3

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

Express $$k$$ as a common fraction.

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

heureka Jun 10, 2019