Question

without solving the quadratic equation find the value of m for which the given equation has real and = roots: x² +2(m-1) x +m(m+5)=0

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}: \\ {\mathrm{x}}^2+2\left(\mathrm{m}-1\right)\mathrm{x}+\mathrm{m}\left(\mathrm{m}+5\right)=0 \\ \mathrm{Now},\mathrm{a}=1;\mathrm{b}=2\left(\mathrm{m}-1\right);\mathrm{c}=\mathrm{m}\left(\mathrm{m}+5\right) \\ \mathrm{Now},\mathrm{D}={\mathrm{b}}^2-\mathrm{ac}=4{\left(\mathrm{m}-1\right)}^2-4\times 1\times \mathrm{m}\left(\mathrm{m}+5\right)=4\left({\mathrm{m}}^2+1-2\mathrm{m}\right)-4{\mathrm{m}}^2-20\mathrm{m} \\ \Rightarrow \mathrm{D}=4{\mathrm{m}}^2+4-8\mathrm{m}-20\mathrm{m}-4{\mathrm{m}}^2=4-28\mathrm{m} \\ \mathrm{Now},\mathrm{for}\mathrm{real}\mathrm{and}\mathrm{equal}\mathrm{roots}: \\ \mathrm{D}=0 \\ \Rightarrow 4-28\mathrm{m}=0 \\ \Rightarrow \mathrm{m}=\frac{4}{28}=\frac{1}{7} \end{gathered}$$