Question

Q12. The value of k for which one root of the equation $x^2-\left(k+1\right)x+k^2+k-8=0$ exceeds 2 and other is less than 2, is
$$\begin{gathered} \left(1\right)(2,3) \\ \left(2\right)(-2,3) \\ \left(3\right)(-3,-2) \\ \left(4\right)(-3,2) \end{gathered}$$

Answer 1

$$\begin{gathered} Rootsof{x}^2-\left(k+1\right)x+k^2+k+8=0isgivenby\frac{(\mathrm{k}+1)+\sqrt{{\left\{-(\mathrm{k}+1)\right\}}^2-4\left(1\right)\left({\mathrm{k}}^2+\mathrm{k}-8\right)}}{2}>2 \\ \Rightarrow (\mathrm{k}+1)+\sqrt{{\left\{-(\mathrm{k}+1)\right\}}^2-4\left(1\right)\left({\mathrm{k}}^2+\mathrm{k}-8\right)}>4 \\ \Rightarrow \sqrt{{\left\{-(\mathrm{k}+1)\right\}}^2-4\left(1\right)\left({\mathrm{k}}^2+\mathrm{k}-8\right)}>4-(\mathrm{k}+1) \\ \Rightarrow \sqrt{{\left\{-(\mathrm{k}+1)\right\}}^2-4\left(1\right)\left({\mathrm{k}}^2+\mathrm{k}-8\right)}>3-\mathrm{k} \\ \Rightarrow {\left\{-(\mathrm{k}+1)\right\}}^2-4\left(1\right)\left({\mathrm{k}}^2+\mathrm{k}-8\right)>{\left(3-\mathrm{k}\right)}^2 \\ \Rightarrow {\mathrm{k}}^2+1+2\mathrm{k}-4{\mathrm{k}}^2-4\mathrm{k}+32>9+{\mathrm{k}}^2-6\mathrm{k} \\ \Rightarrow -4{\mathrm{k}}^2+4\mathrm{k}+24>0 \\ \Rightarrow -{\mathrm{k}}^2+\mathrm{k}+6>0 \\ \Rightarrow {\mathrm{k}}^2-\mathrm{k}-6<0 \\ \Rightarrow (\mathrm{k}+2)(\mathrm{k}-3)<0 \\ \Rightarrow \mathrm{k}\in \left(-2,3\right) \end{gathered}$$
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