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Question Number 106572 by DeepakMahato last updated on 06/Aug/20

Commented by Rasheed.Sindhi last updated on 06/Aug/20

Still another way _(−) Roots(zeros) of ax^2 +bx+c=0 are: x=((−b±(√(b^2 −4ac)))/(2a))=((−b)/(2a))±((√(b^2 −4ac))/(2a)) If the roots are oppsite to each other,then ((−b)/(2a))=0 So ((−(−8k))/(2(4)))=0⇒k=0

$$\:\:\:\:\underset{−} {\:\:\:\:\:\:\:{Still}\:{another}\:{way}\:\:\:\:\:\:\:\:} \\ $$$${Roots}\left({zeros}\right)\:{of}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${are}: \\ $$$$\:{x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}=\frac{−{b}}{\mathrm{2}{a}}\pm\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${If}\:{the}\:{roots}\:{are}\:{oppsite}\:{to}\:{each} \\ $$$${other},{then}\:\frac{−{b}}{\mathrm{2}{a}}=\mathrm{0} \\ $$$$\:{So}\:\:\:\:\:\:\frac{−\left(−\mathrm{8}{k}\right)}{\mathrm{2}\left(\mathrm{4}\right)}=\mathrm{0}\Rightarrow{k}=\mathrm{0} \\ $$

Commented by DeepakMahato last updated on 06/Aug/20

Thank You Sir

Answered by bemath last updated on 06/Aug/20

say the roots of polynome is → { (x_1 ),((x_2 =−x_1 )) :} then by Vieta′s rule ⇒x_1 +x_2 =−(b/a) ((8k)/4) = 0 ⇒ k = 0 then f(x) = 4x^2 −9=(2x+3)(2x−3)

$$\mathrm{say}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{polynome}\:\mathrm{is}\:\rightarrow\begin{cases}{\mathrm{x}_{\mathrm{1}} }\\{\mathrm{x}_{\mathrm{2}} =−\mathrm{x}_{\mathrm{1}} }\end{cases} \\ $$$$\mathrm{then}\:\mathrm{by}\:\mathrm{Vieta}'\mathrm{s}\:\mathrm{rule}\:\Rightarrow\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =−\frac{\mathrm{b}}{\mathrm{a}} \\ $$$$\frac{\mathrm{8k}}{\mathrm{4}}\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{k}\:=\:\mathrm{0}\: \\ $$$$\mathrm{then}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{4x}^{\mathrm{2}} −\mathrm{9}=\left(\mathrm{2x}+\mathrm{3}\right)\left(\mathrm{2x}−\mathrm{3}\right) \\ $$

Commented by DeepakMahato last updated on 06/Aug/20

Thank You Sir��

Answered by Rasheed.Sindhi last updated on 06/Aug/20

AnOther Way _(−) 4(α)^2 −8k(α)−9=0=4(−α)^2 −8k(−α)−9 4α^2 −8kα−9=4α^2 +8kα−9 −8k=8k⇒k=0

$$\:\:\:\:\underset{−} {\:\:\:\:{AnOther}\:{Way}\:\:\:\:} \\ $$$$\mathrm{4}\left(\alpha\right)^{\mathrm{2}} −\mathrm{8}{k}\left(\alpha\right)−\mathrm{9}=\mathrm{0}=\mathrm{4}\left(−\alpha\right)^{\mathrm{2}} −\mathrm{8}{k}\left(−\alpha\right)−\mathrm{9} \\ $$$$\mathrm{4}\alpha^{\mathrm{2}} −\mathrm{8}{k}\alpha−\mathrm{9}=\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{8}{k}\alpha−\mathrm{9} \\ $$$$−\mathrm{8}{k}=\mathrm{8}{k}\Rightarrow{k}=\mathrm{0} \\ $$

Commented by john santu last updated on 06/Aug/20

how to make underline sir?

$$\mathrm{how}\:\mathrm{to}\:\mathrm{make}\:\mathrm{underline}\:\mathrm{sir}? \\ $$

Commented by Rasheed.Sindhi last updated on 06/Aug/20

Commented by Rasheed.Sindhi last updated on 06/Aug/20

1) Select the text 2)Press the shown structure 3)Click minus sign ′−′

$$\left.\mathrm{1}\right)\:{Select}\:{the}\:{text} \\ $$$$\left.\mathrm{2}\right){Press}\:{the}\:{shown}\:{structure} \\ $$$$\left.\mathrm{3}\right){Click}\:{minus}\:{sign}\:'−' \\ $$

Commented by john santu last updated on 06/Aug/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\: \\ $$

Commented by Tinku Tara last updated on 06/Aug/20

In offline editor there is select and underline option.

$$\mathrm{In}\:\mathrm{offline}\:\mathrm{editor}\:\mathrm{there}\:\mathrm{is}\:\mathrm{select} \\ $$$$\mathrm{and}\:\mathrm{underline}\:\mathrm{option}. \\ $$

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