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Question Number 101056 by bemath last updated on 30/Jun/20

$$\mathrm{If}\mathrm{the}\mathrm{equation}4x^2-4(5x+1)+p^2=0$$$$\mathrm{has}\mathrm{one}\mathrm{root}\mathrm{equals}\mathrm{to}\mathrm{two}\mathrm{more}$$$$\mathrm{then}\mathrm{the}\mathrm{other},\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}$$$$p\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Commented by bemath last updated on 30/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Answered by Rasheed.Sindhi last updated on 30/Jun/20

$$\mathbf{Still}\mathbf{another}\mathbf{way}$$$$4x^2-20x+p^2-4=0.......\left(i\right)$$$$Theequationwith\alpha \mathrm{\&}\alpha +2roots$$$$(x-\alpha )(x-\alpha -2)=0$$$$x^2-(2\alpha +2)x+{\alpha }^2+2\alpha =0....\left(ii\right)$$$$Comparingcoefficientsof\left(i\right)\mathrm{\&}\left(ii\right)$$$$\frac{4}{1}=\frac{-20}{-(2\alpha +2)}=\frac{p^2-4}{{\alpha }^2+2\alpha }$$$$8\alpha +8=20\wedge p^2-4=4{\alpha }^2+8\alpha$$$$\alpha =\frac{3}{2}\wedge p^2=4{\left(\frac{3}{2}\right)}^2+8\left(\frac{3}{2}\right)+4$$$$p^2=9+12+4=25$$$$p=\pm 5$$

Commented by bemath last updated on 30/Jun/20

$$\mathrm{waw}...\mathrm{great}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

Answered by Rasheed.Sindhi last updated on 30/Jun/20

$$4x^2-4(5x+1)+p^2=0$$$$4x^2-20x-4+p^2=0$$$$\alpha +(\alpha +2)=\frac{-(-20)}{4}\wedge \alpha (\alpha +2)=\frac{p^2-4}{4}$$$$2\alpha +2=5\Rightarrow \alpha =\frac{3}{2}$$$$\Rightarrow \frac{p^2-4}{4}=\alpha (\alpha +2)=\frac{3}{2}(\frac{3}{2}+2)$$$$\Rightarrow \frac{p^2-4}{4}=\frac{3}{2}\left(\frac{7}{2}\right)$$$$\Rightarrow p^2=21+4\Rightarrow p=\pm 5$$

Answered by Ar Brandon last updated on 30/Jun/20

$$\mathrm{Let}\mathrm{the}\mathrm{roots}\mathrm{be}\alpha \mathrm{and}\alpha +2.\mathrm{Then};$$$$\mathrm{Sum}\mathrm{of}\mathrm{roots}2\alpha +2=\frac{20}{4}=5\Rightarrow \alpha =\frac{3}{2}$$$$\mathrm{Product}\mathrm{of}\mathrm{roots}\alpha (\alpha +2)={\alpha }^2+2\alpha =\frac{{\mathrm{p}}^2-4}{4}$$$$\Rightarrow {\left(\frac{3}{2}\right)}^2+2\left(\frac{3}{2}\right)=\frac{{\mathrm{p}}^2-4}{4}=\frac{21}{4}\Rightarrow \mathrm{p}=\pm 5$$

Answered by Rasheed.Sindhi last updated on 30/Jun/20

$$\mathbf{AnOther}\mathbf{Way}$$$$4x^2-20x+p^2-4=0$$$$\alpha isaroot\left(say\right)$$$$4{\alpha }^2-20\alpha +p^2-4=0.......\left(i\right)$$$$\alpha +2isotherroot$$$$4{(\alpha +2)}^2-20(\alpha +2)+p^2-4=0$$$$4{\alpha }^2+16\alpha +16-20\alpha -40+p^2-4=0$$$$4{\alpha }^2-4\alpha +p^2-28=0.......\left(ii\right)$$$$\left(i\right)-\left(ii\right):-16\alpha +24=0\Rightarrow \alpha =\frac{3}{2}$$$$Now,$$$$\left(i\right)\Rightarrow 4{\left(\frac{3}{2}\right)}^2-20\left(\frac{3}{2}\right)+p^2-4=0$$$$9-30+p^2-4=0\Rightarrow p=\pm 5$$

Answered by Rasheed.Sindhi last updated on 30/Jun/20

$$\mathbf{One}\mathbf{way}\mathbf{more}:\mathrm{A}\mathrm{simple}\mathrm{way}$$$$(Whetheryoulikeitordislike,$$$$anywaythisisalsoaway.)$$$$4x^2-20x+p^2-4=0$$$$x=\frac{5\pm \sqrt{29-p^2}}{2}$$$$\frac{5+\sqrt{29-p^2}}{2}-\frac{5-\sqrt{29-p^2}}{2}=2$$$$\frac{5+\sqrt{29-p^2}-5+\sqrt{29-p^2}}{2}=2$$$$\sqrt{29-p^2}=2$$$$29-p^2=4$$$$p^2=25$$$$p=\pm 5$$

Commented by john santu last updated on 01/Jul/20

$$\mathrm{cooll}$$

Commented by Rasheed.Sindhi last updated on 01/Jul/20

$$\mathcal{T}hanks\mathcal{S}ir!$$

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