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Hi Lindah, The quadratic $(x - r_1)(x - r_2)$ has roots $r_1$ and $r_2$ but so does $2 (x - r_1)(x - r_2)$ and $5 (x - r_1)(x - r_2).$ In fact for any constant $k,$ $k (x - r_1)(x - r_2)$ has roots $r_1$ and $r_2.$ If you know what the constant term is can you find the appropriate value of $k?$ Penny | ||||||||||||
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Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. |