Question

if the range of the values of 'a' for which the roots of the equation x² -2x-a² +1 = 0 lie between the roots of the equation x²-2(a+1)x+a(a+1) =0 is (p,q), then find the value of (q-1/p)

Answer 1

Dear student

$$\begin{gathered} {\mathrm{x}}^2-2\mathrm{x}-{\mathrm{a}}^2+1=0 \\ \mathrm{a}=1,\mathrm{b}=-2,\mathrm{c}=-{\mathrm{a}}^2+1 \\ \mathrm{Using}\mathrm{discriminant}\mathrm{method} \\ \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ \mathrm{x}=\frac{2\pm \sqrt{4-4\times \left(1\right)\times \left(-{\mathrm{a}}^2+1\right)}}{2} \\ \mathrm{x}=\frac{2\pm 2\sqrt{1-1+{\mathrm{a}}^2}}{2} \\ =1\pm \sqrt{{\mathrm{a}}^2} \\ \mathrm{x}=1+\mathrm{a}\mathrm{or}1-\mathrm{a} \\ \mathrm{and}{\mathrm{x}}^2-2\left(\mathrm{a}+1\right)\mathrm{x}+\mathrm{a}\left(\mathrm{a}+1\right)=0 \\ \mathrm{x}=\frac{2\left(\mathrm{a}+1\right)\pm \sqrt{4{\left(\mathrm{a}+1\right)}^2-4\mathrm{a}\left(\mathrm{a}+1\right)}}{2} \\ =\frac{2\left(\mathrm{a}+1\right)\pm \sqrt{4{\left(\mathrm{a}+1\right)}^2-4\mathrm{a}\left(\mathrm{a}+1\right)}}{2} \\ =\left(\mathrm{a}+1\right)\pm \sqrt{{\left(\mathrm{a}+1\right)}^2-\mathrm{a}\left(\mathrm{a}+1\right)} \\ =\left(\mathrm{a}+1\right)\pm \sqrt{{\mathrm{a}}^2+1+2\mathrm{a}-{\mathrm{a}}^2-\mathrm{a}} \\ =\left(\mathrm{a}+1\right)\pm \sqrt{1+\mathrm{a}} \\ \mathrm{As}\mathrm{per}\mathrm{the}\mathrm{question}\left(\mathrm{a}+1\right)-\sqrt{1+\mathrm{a}}<1-\mathrm{a}\mathrm{and}\left(\mathrm{a}+1\right)+\sqrt{1+\mathrm{a}}>1+\mathrm{a} \end{gathered}$$
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