Question

If a square of a number is multiplied with the successor of that number then the result is 132 times the number (which you squared). what could be the numbers ?

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{be}\mathrm{x}\mathrm{and}\mathrm{its}\mathrm{successor}\mathrm{be}\mathrm{x}+1 \\ \mathrm{So},\mathrm{A}.\mathrm{T}.\mathrm{Q} \\ {\mathrm{x}}^2\left(\mathrm{x}+1\right)=132\mathrm{x} \\ {\mathrm{x}}^3+{\mathrm{x}}^2-132\mathrm{x}=0 \\ \mathrm{x}\left({\mathrm{x}}^2+\mathrm{x}-132\right)=0 \\ \mathrm{x}\left[{\mathrm{x}}^2+12\mathrm{x}-11\mathrm{x}-132\right]=0 \\ \mathrm{x}\left[\mathrm{x}\left(\mathrm{x}+12\right)-11\left(\mathrm{x}+12\right)\right]=0 \\ \mathrm{x}\left(\mathrm{x}-11\left)\right(\mathrm{x}+12\right)=0 \\ \mathrm{x}=0,\mathrm{x}-11=0,\mathrm{x}+12=0 \\ \mathrm{x}=0,\mathrm{x}=11,\mathrm{x}=-12 \\ \mathrm{So},\mathrm{x}=11\mathrm{and}\mathrm{x}=0\mathrm{and}\mathrm{x}=-12(\mathrm{Not}\mathrm{possible}) \\ \mathrm{So},\mathrm{the}\mathrm{numbers}\mathrm{will}\mathrm{be}11\mathrm{and}12 \end{gathered}$$
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