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Question Number 204461 by MATHEMATICSAM last updated on 18/Feb/24

Solve for x (x − 12)(x − 13) = ((34)/(33^2 ))

$$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$$\left({x}\:−\:\mathrm{12}\right)\left({x}\:−\:\mathrm{13}\right)\:=\:\frac{\mathrm{34}}{\mathrm{33}^{\mathrm{2}} } \\ $$

Answered by TonyCWX08 last updated on 18/Feb/24

x^2 −25x+156 = ((34)/(1089)) x^2 −25 = ((34)/(1089))−156 x^2 −25+(((25)/2))^2 =((−164859)/(1057))+((625)/4) (x−((25)/2))^2 =((304)/(1081)) x−((25)/2)=±(√((304)/(1081))) x=((25)/2)±(√((304)/(1081))) x_1 =((395)/(33)) x_2 =((430)/(33))

$${x}^{\mathrm{2}} −\mathrm{25}{x}+\mathrm{156}\:=\:\frac{\mathrm{34}}{\mathrm{1089}} \\ $$$${x}^{\mathrm{2}} −\mathrm{25}\:=\:\frac{\mathrm{34}}{\mathrm{1089}}−\mathrm{156} \\ $$$${x}^{\mathrm{2}} −\mathrm{25}+\left(\frac{\mathrm{25}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{−\mathrm{164859}}{\mathrm{1057}}+\frac{\mathrm{625}}{\mathrm{4}} \\ $$$$\left({x}−\frac{\mathrm{25}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{304}}{\mathrm{1081}} \\ $$$${x}−\frac{\mathrm{25}}{\mathrm{2}}=\pm\sqrt{\frac{\mathrm{304}}{\mathrm{1081}}} \\ $$$${x}=\frac{\mathrm{25}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{304}}{\mathrm{1081}}} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{395}}{\mathrm{33}}\:{x}_{\mathrm{2}} =\frac{\mathrm{430}}{\mathrm{33}} \\ $$

Answered by som(math1967) last updated on 18/Feb/24

let 33=a⇒34=a+1 (x−12)(x−13)=((a+1)/a^2 ) ⇒a^2 (x−12)(x−13)−a−1=0 ⇒a^2 (x−12)(x−13)+(x−13)a −(x−12)a−1=0 ⇒a(x−13)(ax−12a+1) −1(ax−12a+1)=0 ⇒(ax−12a+1)(ax−13a−1)=0 ∴x=((12a−1)/a)=12−(1/a)=12−(1/(33)) x=11((32)/(33)) or x=((13a+1)/a)=13+(1/a)=13(1/(33))

$${let}\:\mathrm{33}={a}\Rightarrow\mathrm{34}={a}+\mathrm{1} \\ $$$$\left({x}−\mathrm{12}\right)\left({x}−\mathrm{13}\right)=\frac{{a}+\mathrm{1}}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{a}^{\mathrm{2}} \left({x}−\mathrm{12}\right)\left({x}−\mathrm{13}\right)−{a}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} \left({x}−\mathrm{12}\right)\left({x}−\mathrm{13}\right)+\left({x}−\mathrm{13}\right){a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left({x}−\mathrm{12}\right){a}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}\left({x}−\mathrm{13}\right)\left({ax}−\mathrm{12}{a}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:−\mathrm{1}\left({ax}−\mathrm{12}{a}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({ax}−\mathrm{12}{a}+\mathrm{1}\right)\left({ax}−\mathrm{13}{a}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\therefore{x}=\frac{\mathrm{12}{a}−\mathrm{1}}{{a}}=\mathrm{12}−\frac{\mathrm{1}}{{a}}=\mathrm{12}−\frac{\mathrm{1}}{\mathrm{33}} \\ $$$${x}=\mathrm{11}\frac{\mathrm{32}}{\mathrm{33}} \\ $$$${or}\:{x}=\frac{\mathrm{13}{a}+\mathrm{1}}{{a}}=\mathrm{13}+\frac{\mathrm{1}}{{a}}=\mathrm{13}\frac{\mathrm{1}}{\mathrm{33}} \\ $$

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