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Question Number 30990 by rahul 19 last updated on 01/Mar/18

Answered by MJS last updated on 01/Mar/18

$$-2x^2+kx+(k^2+5)=0$$$$x_1=\frac{k}{4}-\frac{\sqrt{9k^2+40}}{4}$$$$x_2=\frac{k}{4}+\frac{\sqrt{9k^2+40}}{4}$$$$x_1=0,x_1=2,x_2=0\Rightarrow \mathrm{no}\mathrm{solution}\mathrm{in}\mathbb{R}$$$$x_2=2\Rightarrow k_1=-3,k_2=1$$$$\Rightarrow -3<k<1$$$$a+10b=7$$

Commented by rahul 19 last updated on 01/Mar/18

$$howugot{x}_1=0,x_1=2,x_2=0$$$$x_2=2\Rightarrow k_1=-3....$$$$plzexplainallthisclearly.$$

Commented by MJS last updated on 01/Mar/18

$$\mathrm{ok}\mathrm{will}\mathrm{do}\mathrm{tomorrow}\mathrm{morning},{\mathrm{it}}^{\prime }\mathrm{s}10:30\mathrm{p}.\mathrm{m}.$$$$\mathrm{over}\mathrm{here}...$$

Commented by rahul 19 last updated on 03/Mar/18

$$??$$

Commented by MJS last updated on 04/Mar/18

$$\mathrm{sorry},{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{late}...$$$$\mathrm{I}\mathrm{set}x\mathrm{on}\mathrm{the}\mathrm{borders},\mathrm{and}\mathrm{solved}$$$$\mathrm{the}\mathrm{equations}\mathrm{for}{x}_1,x_2$$$$(\mathrm{these}\mathrm{show}\mathrm{that}f\left(x\right)\mathrm{always}\mathrm{has}$$$$2\mathrm{distinct}\mathrm{zeroes},\mathrm{because}\mathrm{the}\mathrm{root}$$$$\mathrm{is}\in {\mathbb{R}}^{+}\mathrm{for}k\in \mathbb{R})$$$$$$$$\mathrm{now}\mathrm{I}\mathrm{see}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easier}\mathrm{to}\mathrm{set}x=0\mathrm{and}$$$$x=2\mathrm{in}\mathrm{the}\mathrm{original}\mathrm{funktion}\mathrm{and}$$$$\mathrm{find}\mathrm{the}\mathrm{zeroes}\mathrm{for}k$$$$x=0\Rightarrow k^2+5=0\Rightarrow \mathrm{no}\mathrm{solution}$$$$x=2\Rightarrow k^2+2k-3=0\Rightarrow$$$$\Rightarrow k_1=-3,k_2=1$$$$\mathrm{which}\mathrm{means}\mathrm{that}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{zeroes}$$$$\mathrm{for}x\mathrm{is}\mathrm{on}\mathrm{the}\mathrm{border}\mathrm{of}\mathrm{the}\mathrm{given}$$$$\mathrm{interval}\mathrm{when}k\mathrm{is}-3\mathrm{or}1$$$$\mathrm{the}\mathrm{other}\mathrm{zero}\mathrm{is}\mathrm{negative}(\mathrm{easy}\mathrm{to}$$$$\mathrm{check}\mathrm{with}\mathrm{the}{x}_1-\mathrm{equation})$$

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