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Question Number 188327 by HeferH last updated on 28/Feb/23

if the roots of 2x^2 −xn = 2x + m is 5, then find : 4n + m − 5

$${if}\:{the}\:{roots}\:{of}\:\:\mathrm{2}{x}^{\mathrm{2}} \:−{xn}\:=\:\mathrm{2}{x}\:+\:{m}\:\:{is}\:\mathrm{5}, \\ $$$$\:{then}\:{find}\::\:\mathrm{4}{n}\:+\:{m}\:−\:\mathrm{5}\: \\ $$$$\: \\ $$

Answered by Rasheed.Sindhi last updated on 28/Feb/23

2x^2 −xn = 2x + m 2x^2 −xn − 2x − m=0 2x^2 −(n+2)x−m=0 Say α , β are roots Given α=β=5 α+β=((n+2)/2)=5+5=10⇒n=18 αβ=−(m/2)=5×5=25⇒m=−50 4n + m − 5=4(18)+(−50)−5=17

$$\mathrm{2}{x}^{\mathrm{2}} \:−{xn}\:=\:\mathrm{2}{x}\:+\:{m} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:−{xn}\:−\:\mathrm{2}{x}\:−\:{m}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\left({n}+\mathrm{2}\right){x}−{m}=\mathrm{0} \\ $$$${Say}\:\alpha\:,\:\beta\:{are}\:{roots} \\ $$$${Given}\:\alpha=\beta=\mathrm{5} \\ $$$$\alpha+\beta=\frac{{n}+\mathrm{2}}{\mathrm{2}}=\mathrm{5}+\mathrm{5}=\mathrm{10}\Rightarrow{n}=\mathrm{18} \\ $$$$\:\alpha\beta=−\frac{{m}}{\mathrm{2}}=\mathrm{5}×\mathrm{5}=\mathrm{25}\Rightarrow{m}=−\mathrm{50} \\ $$$$ \\ $$$$\mathrm{4}{n}\:+\:{m}\:−\:\mathrm{5}=\mathrm{4}\left(\mathrm{18}\right)+\left(−\mathrm{50}\right)−\mathrm{5}=\mathrm{17} \\ $$$$ \\ $$

Commented by HeferH last updated on 28/Feb/23

Nice solution! I forgor about those properties of roots

$${Nice}\:{solution}!\:{I}\:{forgor}\:{about}\:{those}\:{properties} \\ $$$$\:{of}\:{roots} \\ $$

Answered by Rasheed.Sindhi last updated on 28/Feb/23

Another way... 2x^2 −xn = 2x + m 2x^2 −xn−2x− m=0 2x^2 −(n+2)x−m=0....(i) The equation having roots 5 , 5 is (x−5)^2 =0 x^2 −10x+25=0.........(ii) Comparing coefficients of (i) & (ii) (2/1)=((−(n+2))/(−10))=(( −m )/(25)) n+2=20 ∧ −m=50 n=18 ∧ m=−50 ▶ 4n + m − 5=4(18)+(−50)−5=17

$${Another}\:{way}... \\ $$$$\:\mathrm{2}{x}^{\mathrm{2}} \:−{xn}\:=\:\mathrm{2}{x}\:+\:{m} \\ $$$$\:\mathrm{2}{x}^{\mathrm{2}} \:−{xn}−\mathrm{2}{x}−\:{m}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\left({n}+\mathrm{2}\right){x}−{m}=\mathrm{0}....\left({i}\right) \\ $$$${The}\:{equation}\:{having}\:{roots}\:\mathrm{5}\:,\:\mathrm{5}\:{is} \\ $$$$\:\:\:\:\left({x}−\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{25}=\mathrm{0}.........\left({ii}\right) \\ $$$${Comparing}\:{coefficients}\:{of}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{1}}=\frac{−\left({n}+\mathrm{2}\right)}{−\mathrm{10}}=\frac{\:−{m}\:}{\mathrm{25}} \\ $$$${n}+\mathrm{2}=\mathrm{20}\:\wedge\:−{m}=\mathrm{50} \\ $$$$\:\:\:{n}=\mathrm{18}\:\wedge\:{m}=−\mathrm{50} \\ $$$$\blacktriangleright\:\mathrm{4}{n}\:+\:{m}\:−\:\mathrm{5}=\mathrm{4}\left(\mathrm{18}\right)+\left(−\mathrm{50}\right)−\mathrm{5}=\mathrm{17} \\ $$

Answered by Rasheed.Sindhi last updated on 28/Feb/23

2x^2 −(n+2)x−m=0 • ∵5 is root ∴2(5)^2 −(n+2)(5)−m=0 50−5n−10−m=0 m+5n=40..........(i) •(d/dx)(2x^2 −(n+2)x−m)_(n=5) =(d/dx)(0)_(x=5) 4x−(n+2)]_(x=5) =0 ⇒4(5)−n−2=0 ⇒n=18 (i)⇒m+5(18)=40⇒m=−50 ▶4n + m − 5=4(18)+(−50)−5=17

$$\mathrm{2}{x}^{\mathrm{2}} −\left({n}+\mathrm{2}\right){x}−{m}=\mathrm{0} \\ $$$$\bullet\:\because\mathrm{5}\:{is}\:{root} \\ $$$$\:\:\:\therefore\mathrm{2}\left(\mathrm{5}\right)^{\mathrm{2}} −\left({n}+\mathrm{2}\right)\left(\mathrm{5}\right)−{m}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{50}−\mathrm{5}{n}−\mathrm{10}−{m}=\mathrm{0} \\ $$$$\:\:\:\:{m}+\mathrm{5}{n}=\mathrm{40}..........\left({i}\right) \\ $$$$\bullet\frac{{d}}{{dx}}\left(\mathrm{2}{x}^{\mathrm{2}} −\left({n}+\mathrm{2}\right){x}−{m}\right)_{{n}=\mathrm{5}} =\frac{{d}}{{dx}}\left(\mathrm{0}\right)_{{x}=\mathrm{5}} \\ $$$$\left.\:\:\mathrm{4}{x}−\left({n}+\mathrm{2}\right)\right]_{{x}=\mathrm{5}} =\mathrm{0} \\ $$$$\:\:\Rightarrow\mathrm{4}\left(\mathrm{5}\right)−{n}−\mathrm{2}=\mathrm{0} \\ $$$$\:\:\Rightarrow{n}=\mathrm{18} \\ $$$$\:\:\:\left({i}\right)\Rightarrow{m}+\mathrm{5}\left(\mathrm{18}\right)=\mathrm{40}\Rightarrow{m}=−\mathrm{50} \\ $$$$\blacktriangleright\mathrm{4}{n}\:+\:{m}\:−\:\mathrm{5}=\mathrm{4}\left(\mathrm{18}\right)+\left(−\mathrm{50}\right)−\mathrm{5}=\mathrm{17}\: \\ $$

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