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Question Number 158097 by zainaltanjung last updated on 31/Oct/21

Answered by Rasheed.Sindhi last updated on 31/Oct/21

x^2 +px+12:Roots α , 4 product of roots:4α=12⇒α=3 (other root) sum of roots:4+3=−p⇒p=−7 x^2 +px+q=0⇒x^2 −7x+q=0 Roots:β &β sum of roots:2β=−7⇒β=−(7/2) product of roots:β^2 =q⇒q=(−(7/2))^2 =((49)/4)

$${x}^{\mathrm{2}} +{px}+\mathrm{12}:{Roots}\:\alpha\:,\:\mathrm{4} \\ $$$$\mathrm{product}\:{of}\:{roots}:\mathrm{4}\alpha=\mathrm{12}\Rightarrow\alpha=\mathrm{3}\:\left({other}\:{root}\right) \\ $$$${sum}\:{of}\:{roots}:\mathrm{4}+\mathrm{3}=−\mathrm{p}\Rightarrow\mathrm{p}=−\mathrm{7} \\ $$$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} −\mathrm{7}{x}+{q}=\mathrm{0} \\ $$$${Roots}:\beta\:\&\beta \\ $$$${sum}\:{of}\:{roots}:\mathrm{2}\beta=−\mathrm{7}\Rightarrow\beta=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${product}\:{of}\:{roots}:\beta^{\mathrm{2}} ={q}\Rightarrow{q}=\left(−\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{49}}{\mathrm{4}} \\ $$

Answered by Rasheed.Sindhi last updated on 31/Oct/21

5x^2 +13x+k=0 Roots: α,(1/α) product of roots: α×(1/α)=k⇒k=1

$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{13}{x}+{k}=\mathrm{0} \\ $$$${Roots}:\:\alpha,\frac{\mathrm{1}}{\alpha} \\ $$$${product}\:{of}\:{roots}:\:\alpha×\frac{\mathrm{1}}{\alpha}={k}\Rightarrow{k}=\mathrm{1} \\ $$

Commented by otchereabdullai@gmail.com last updated on 31/Oct/21

nice one!

$$\mathrm{nice}\:\mathrm{one}! \\ $$

Commented by Rasheed.Sindhi last updated on 31/Oct/21

ThanX abdullai!

$$\mathcal{T}{han}\mathcal{X}\:{abdullai}! \\ $$

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