Question

The equation of x²+2x=k(x-1) has non-zero roots where the difference between the root is 2, find the value of each root and the value of k

Answer 1

`k=8.\ x_1=2.\ x_2=4`

Explanation:

`\mathrm{We\ have,}\\x^2+2x=k(x-1)\\\mathrm{or,\ }x^2+2x=kx-k\\\mathrm{or,\ }x^2+2x-kx+k=0\\\mathrm{or,\ }x^2+(2-k)x+k=0\\`

`\mathrm{Let}\ p\ \mathrm{and}\ p-2\ \mathrm{be\ the\ roots\ of\ the\ equation.}\\\mathrm{Then,}\\p+p-2=-(2-k)\\\mathrm{or,\ }2p-2=k-2\\\mathrm{or,\ }k=2p.....(1)\\\mathrm{Also,}\\p(p-2)=k\\\\mathrm{From\ equation(1),}\\p(p-2)=2p\\\mathrm{or,\ }p^2-2p=2p\\\mathrm{or,\ }p^2-4p=0\\\mathrm{or,\ }p(p-4)=0\\\mathrm{i.e.\ }p=0\ \mathrm{or}\ p=4`

`\mathrm{When\ }p=0,\ k=2(0)=0.\\\mathrm{When\ }p=4.\ k=2(4)=8.\\\mathrm{Here,}\ k=0\ \mathrm{is\ neglected\ because\ on\ taking\ k=0, \ quadratic\ equation\ }\\\mathrm{becomes\ }x^2+2x=0\ \mathrm{whose\ one\ root\ is\ 0}.`

`\mathrm{When\ }k=8,\ \mathrm{the\ quadratic\ equation\ becomes:}\\x^2+(2-8)x+8=0\\\mathrm{or,\ }x^2-6x+8=0\\\mathrm{or,\ }x^2-2x-4x+8=0\\\mathrm{or,\ }x(x-2)-4(x-2)=0\\\mathrm{or,\ }(x-2)(x-4)=0\\\mathrm{i.e.\ }x=2\ \mathrm{or}\ 4.\\\mathrm{So\ the\ value\ of\ each\ roots\ are\ 2\ and\ 4\ and\ the\ value\ of\ k\ is\ 8.}`