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Question Number 147011 by mathdanisur last updated on 17/Jul/21

2x - (√(2x - 3)) - 9 = 0 if there′s a solution to equation, find 4a + 3 = ?

$$\mathrm{2}{x}\:-\:\sqrt{\mathrm{2}{x}\:-\:\mathrm{3}}\:-\:\mathrm{9}\:=\:\mathrm{0} \\ $$$${if}\:{there}'{s}\:\boldsymbol{{a}}\:{solution}\:{to}\:{equation}, \\ $$$${find}\:\:\mathrm{4}\boldsymbol{{a}}\:+\:\mathrm{3}\:=\:? \\ $$

Commented by 7770 last updated on 17/Jul/21

4x^2 −36x+81−2x+3=0 4x^2 −38x+84=0 x_1 =6 x_2 =3.5 ∴ 4(3.5)+3=17 4(6)+3=27

$$\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{36}\boldsymbol{{x}}+\mathrm{81}−\mathrm{2}\boldsymbol{{x}}+\mathrm{3}=\mathrm{0} \\ $$$$\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{38}\boldsymbol{{x}}+\mathrm{84}=\mathrm{0} \\ $$$$\:\boldsymbol{{x}}_{\mathrm{1}} =\mathrm{6}\:\:\:\:\:\:\boldsymbol{{x}}_{\mathrm{2}} =\mathrm{3}.\mathrm{5} \\ $$$$\:\therefore\:\mathrm{4}\left(\mathrm{3}.\mathrm{5}\right)+\mathrm{3}=\mathrm{17} \\ $$$$\:\:\:\:\:\mathrm{4}\left(\mathrm{6}\right)+\mathrm{3}=\mathrm{27} \\ $$

Commented by otchereabdullai@gmail.com last updated on 18/Jul/21

nice question

$$\mathrm{nice}\:\mathrm{question} \\ $$

Commented by mathdanisur last updated on 18/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

Answered by gsk2684 last updated on 17/Jul/21

((√(2a−3)))^2 −(√(2a−3))−6=0 t^2 −t−6=0 (t−3)(t+2)=0 t=3 (√(2a−3))=3 2a=12 4a+3=27

$$\left(\sqrt{\mathrm{2}{a}−\mathrm{3}}\right)^{\mathrm{2}} −\sqrt{\mathrm{2}{a}−\mathrm{3}}−\mathrm{6}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −{t}−\mathrm{6}=\mathrm{0} \\ $$$$\left({t}−\mathrm{3}\right)\left({t}+\mathrm{2}\right)=\mathrm{0} \\ $$$${t}=\mathrm{3}\: \\ $$$$\sqrt{\mathrm{2}{a}−\mathrm{3}}=\mathrm{3} \\ $$$$\mathrm{2}{a}=\mathrm{12} \\ $$$$\mathrm{4}{a}+\mathrm{3}=\mathrm{27} \\ $$

Commented by mathdanisur last updated on 17/Jul/21

thankyou Ser, but answer 17

$${thankyou}\:{Ser},\:{but}\:{answer}\:\mathrm{17} \\ $$

Commented by gsk2684 last updated on 17/Jul/21

if so then a=((17−3)/4)=((14)/4)=(7/2) 2a−(√(2a−3))−9 7−2−9 ≠0

$${if}\:{so}\:{then}\:{a}=\frac{\mathrm{17}−\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{14}}{\mathrm{4}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\mathrm{2}{a}−\sqrt{\mathrm{2}{a}−\mathrm{3}}−\mathrm{9} \\ $$$$\mathrm{7}−\mathrm{2}−\mathrm{9} \\ $$$$\neq\mathrm{0} \\ $$

Commented by mathdanisur last updated on 18/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

Answered by Rasheed.Sindhi last updated on 17/Jul/21

2x - (√(2x - 3)) - 9 = 0 4a+3=? a: solution to the equation_(−) 4x^2 −36x+81=2x−3 4x^2 −38x+84=0 2x^2 −19x+42=0 (x−6)(2x−7)=0 x=6 ∨ x=7/2 (Extaneous root) a=6 4a+3=4(6)+3=27 ( If x=7/2 were satisfied then 4a+3=4((7/2))+3=17 )

$$\mathrm{2}{x}\:-\:\sqrt{\mathrm{2}{x}\:-\:\mathrm{3}}\:-\:\mathrm{9}\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\underset{−} {\mathrm{4}\boldsymbol{{a}}+\mathrm{3}=?\:\boldsymbol{{a}}:\:{solution}\:{to}\:{the}\:{equation}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{36}{x}+\mathrm{81}=\mathrm{2}{x}−\mathrm{3} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{38}{x}+\mathrm{84}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{19}{x}+\mathrm{42}=\mathrm{0} \\ $$$$\left({x}−\mathrm{6}\right)\left(\mathrm{2}{x}−\mathrm{7}\right)=\mathrm{0} \\ $$$${x}=\mathrm{6}\:\vee\:{x}=\mathrm{7}/\mathrm{2}\:\left(\mathcal{E}{xtaneous}\:{root}\right) \\ $$$$\boldsymbol{{a}}=\mathrm{6} \\ $$$$\mathrm{4}\boldsymbol{{a}}+\mathrm{3}=\mathrm{4}\left(\mathrm{6}\right)+\mathrm{3}=\mathrm{27} \\ $$$$\left(\:{If}\:{x}=\mathrm{7}/\mathrm{2}\:\boldsymbol{{were}}\:\boldsymbol{{satisfied}}\:\right. \\ $$$$\left.{then}\:\mathrm{4}\boldsymbol{{a}}+\mathrm{3}=\mathrm{4}\left(\frac{\mathrm{7}}{\mathrm{2}}\right)+\mathrm{3}=\mathrm{17}\:\right) \\ $$

Commented by mathdanisur last updated on 17/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

Commented by mathdanisur last updated on 18/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

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