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For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution? Express your answer as a common fraction.

May 4, 2018

#1
+3

For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution?

$$\begin{array}{|rcll|} \hline (x+4)(x+1) &=& m + 2x \\ x^2+5x+4 &=& m+2x \\ x^2+3x+4-m &=& 0 \\\\ x &=& \dfrac{-3\pm \sqrt{9-4(4-m)} }{2} \\ &=& \dfrac{-3\pm \sqrt{9-16+4m} }{2} \\ &=& \dfrac{-3\pm \sqrt{4m-7} }{2} \\\\ && \text{one real solution:  \sqrt{4m-7}=0} \\ \sqrt{4m-7} &=& 0 \\ 4m-7 &=& 0 \\ 4m &=& 7 \\ \mathbf{ m } & \mathbf{=} & \mathbf{ \dfrac{7}{4} } \\ \hline \end{array}$$ May 4, 2018

#1
+3

For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution?

$$\begin{array}{|rcll|} \hline (x+4)(x+1) &=& m + 2x \\ x^2+5x+4 &=& m+2x \\ x^2+3x+4-m &=& 0 \\\\ x &=& \dfrac{-3\pm \sqrt{9-4(4-m)} }{2} \\ &=& \dfrac{-3\pm \sqrt{9-16+4m} }{2} \\ &=& \dfrac{-3\pm \sqrt{4m-7} }{2} \\\\ && \text{one real solution:  \sqrt{4m-7}=0} \\ \sqrt{4m-7} &=& 0 \\ 4m-7 &=& 0 \\ 4m &=& 7 \\ \mathbf{ m } & \mathbf{=} & \mathbf{ \dfrac{7}{4} } \\ \hline \end{array}$$ 