Question

solve the following quadratic equation by the method of completing squares and verify the solution by quadratc formula:5x2+6x+9=0

Answer 1

Hi,

$$\begin{gathered} 5{\mathrm{x}}^2+6\mathrm{x}+9=0 \\ \Rightarrow 5\left[{\mathrm{x}}^2+\frac{6\mathrm{x}}{5}+\frac{9}{5}\right]=0 \\ \Rightarrow {\mathrm{x}}^2+\frac{6\mathrm{x}}{5}+\frac{9}{5}=0 \\ \Rightarrow {\mathrm{x}}^2+\frac{6\mathrm{x}}{5}+\frac{9}{25}+\frac{9}{5}-\frac{9}{25}=0 \\ \Rightarrow {\mathrm{x}}^2+\frac{6\mathrm{x}}{5}+\frac{9}{25}=\frac{9}{25}-\frac{9}{5} \\ \Rightarrow {\left(\mathrm{x}+\frac{3}{5}\right)}^2=\frac{9-45}{25} \\ \Rightarrow {\left(\mathrm{x}+\frac{3}{5}\right)}^2=-\frac{36}{25} \\ \mathrm{The}\mathrm{above}\mathrm{equation}\mathrm{can}\text{'}\mathrm{t}\mathrm{be}\mathrm{true}\mathrm{for}\mathrm{any}\mathrm{real}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{as}{\left(\mathrm{x}+\frac{3}{5}\right)}^2\mathrm{can}\text{'}\mathrm{t}\mathrm{be}\mathrm{negative}\mathrm{for}\mathrm{any}\mathrm{real}\mathrm{x}. \\ \mathrm{So},\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{has}\mathrm{no}\mathrm{real}\mathrm{roots}. \end{gathered}$$

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