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

If the equation 4x^2 −4(5x+1)+p^2 =0 has one root equals to two more then the other, then the value of p is equal to ___

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

Commented by bemath last updated on 30/Jun/20

thank you both

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

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

Still another way 4x^2 −20x+p^2 −4=0.......(i) The equation with α & α+2 roots (x−α)(x−α−2)=0 x^2 −(2α+2)x+α^2 +2α=0....(ii) Comparing coefficients of (i)&(ii) (4/1)=((−20)/(−(2α+2)))=((p^2 −4)/(α^2 +2α)) 8α+8=20 ∧ p^2 −4=4α^2 +8α α=(3/2) ∧ p^2 =4((3/2))^2 +8((3/2))+4 p^2 =9+12+4=25 p=±5

$$\boldsymbol{\mathrm{Still}}\:\boldsymbol{\mathrm{another}}\:\boldsymbol{\mathrm{way}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{20}{x}+{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{0}.......\left({i}\right) \\ $$$${The}\:{equation}\:{with}\:\alpha\:\&\:\alpha+\mathrm{2}\:{roots} \\ $$$$\:\:\:\left({x}−\alpha\right)\left({x}−\alpha−\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:{x}^{\mathrm{2}} −\left(\mathrm{2}\alpha+\mathrm{2}\right){x}+\alpha^{\mathrm{2}} +\mathrm{2}\alpha=\mathrm{0}....\left({ii}\right) \\ $$$${Comparing}\:{coefficients}\:{of}\:\left({i}\right)\&\left({ii}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{1}}=\frac{−\mathrm{20}}{−\left(\mathrm{2}\alpha+\mathrm{2}\right)}=\frac{{p}^{\mathrm{2}} −\mathrm{4}}{\alpha^{\mathrm{2}} +\mathrm{2}\alpha} \\ $$$$\mathrm{8}\alpha+\mathrm{8}=\mathrm{20}\:\wedge\:{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{8}\alpha \\ $$$$\alpha=\frac{\mathrm{3}}{\mathrm{2}}\:\:\wedge\:{p}^{\mathrm{2}} =\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{8}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{4} \\ $$$$\:\:\:\:\:\:\:{p}^{\mathrm{2}} =\mathrm{9}+\mathrm{12}+\mathrm{4}=\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}=\pm\mathrm{5} \\ $$

Commented by bemath last updated on 30/Jun/20

waw...great sir. thank you

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

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 α+(α+2)=((−(−20))/4) ∧ α(α+2)=((p^2 −4)/4) 2α+2=5⇒α=(3/2) ⇒((p^2 −4)/4)=α(α+2)=(3/2)((3/2)+2) ⇒((p^2 −4)/4)=(3/2)((7/2)) ⇒p^2 =21+4⇒p=±5

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

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

Let the roots be α and α+2. Then; Sum of roots 2α+2=((20)/4)=5⇒α=(3/2) Product of roots α(α+2)=α^2 +2α=((p^2 −4)/4) ⇒((3/2))^2 +2((3/2))=((p^2 −4)/4)=((21)/4)⇒p=±5

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

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

AnOther Way 4x^2 −20x+p^2 −4=0 α is a root (say) 4α^2 −20α+p^2 −4=0.......(i) α+2 is other root 4(α+2)^2 −20(α+2)+p^2 −4=0 4α^2 +16α+16−20α−40+p^2 −4=0 4α^2 −4α+p^2 −28=0.......(ii) (i)−(ii):−16α+24=0⇒α=(3/2) Now , (i)⇒4((3/2))^2 −20((3/2))+p^2 −4=0 9−30+p^2 −4=0⇒p=±5

$$\boldsymbol{\mathrm{AnOther}}\:\boldsymbol{\mathrm{Way}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{20}{x}+{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\alpha\:{is}\:{a}\:{root}\:\left({say}\right) \\ $$$$\mathrm{4}\alpha^{\mathrm{2}} −\mathrm{20}\alpha+{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{0}.......\left({i}\right) \\ $$$$\alpha+\mathrm{2}\:{is}\:{other}\:{root} \\ $$$$\mathrm{4}\left(\alpha+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{20}\left(\alpha+\mathrm{2}\right)+{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{4}\alpha^{\mathrm{2}} +\mathrm{16}\alpha+\mathrm{16}−\mathrm{20}\alpha−\mathrm{40}+{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{4}\alpha^{\mathrm{2}} −\mathrm{4}\alpha+{p}^{\mathrm{2}} −\mathrm{28}=\mathrm{0}.......\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right):−\mathrm{16}\alpha+\mathrm{24}=\mathrm{0}\Rightarrow\alpha=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${Now}\:, \\ $$$$\left({i}\right)\Rightarrow\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{20}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{9}−\mathrm{30}+{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{0}\Rightarrow{p}=\pm\mathrm{5} \\ $$

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

One way more:A simple way (Whether you like it or dislike, anyway this is also a way.) 4x^2 −20x+p^2 −4=0 x=((5±(√(29−p^2 )))/2) ((5+(√(29−p^2 )))/2)−((5−(√(29−p^2 )))/2)=2 ((5+(√(29−p^2 ))−5+(√(29−p^2 )))/2)=2 (√(29−p^2 ))=2 29−p^2 =4 p^2 =25 p=±5

$$\boldsymbol{\mathrm{One}}\:\boldsymbol{\mathrm{way}}\:\boldsymbol{\mathrm{more}}:\mathrm{A}\:\mathrm{simple}\:\mathrm{way} \\ $$$$\left({Whether}\:{you}\:{like}\:{it}\:{or}\:{dislike},\right. \\ $$$$\left.\:\:\:\:{anyway}\:{this}\:{is}\:{also}\:{a}\:{way}.\right)\: \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{20}{x}+{p}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{5}\pm\sqrt{\mathrm{29}−{p}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\frac{\mathrm{5}+\sqrt{\mathrm{29}−{p}^{\mathrm{2}} }}{\mathrm{2}}−\frac{\mathrm{5}−\sqrt{\mathrm{29}−{p}^{\mathrm{2}} }}{\mathrm{2}}=\mathrm{2} \\ $$$$\frac{\mathrm{5}+\sqrt{\mathrm{29}−{p}^{\mathrm{2}} }−\mathrm{5}+\sqrt{\mathrm{29}−{p}^{\mathrm{2}} }}{\mathrm{2}}=\mathrm{2} \\ $$$$\sqrt{\mathrm{29}−{p}^{\mathrm{2}} }=\mathrm{2} \\ $$$$\mathrm{29}−{p}^{\mathrm{2}} =\mathrm{4} \\ $$$${p}^{\mathrm{2}} =\mathrm{25} \\ $$$${p}=\pm\mathrm{5} \\ $$

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

cooll

$$\mathrm{cooll} \\ $$

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

Thanks Sir!

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

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