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Question Number 161369 by mnjuly1970 last updated on 17/Dec/21

       Determine the value of the following         proposition . (  True  or  False )           ∃ x ∈ R ;   determinant ((( 1+2x),( 2x),(2x)),((  2x),( 1+2x),( 2x )),((  2x),( 2x),(1 +2x)))= x^( 3) + 8x−2                      −−−−−−−−−

$$ \\ $$$$\:\:\:\:\:\mathrm{D}{etermine}\:{the}\:{value}\:{of}\:{the}\:{following} \\ $$$$\:\:\:\:\:\:\:{proposition}\:.\:\left(\:\:\mathrm{T}{rue}\:\:{or}\:\:\mathrm{F}{alse}\:\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\exists\:{x}\:\in\:\mathbb{R}\:;\:\:\begin{vmatrix}{\:\mathrm{1}+\mathrm{2}{x}}&{\:\mathrm{2}{x}}&{\mathrm{2}{x}}\\{\:\:\mathrm{2}{x}}&{\:\mathrm{1}+\mathrm{2}{x}}&{\:\mathrm{2}{x}\:}\\{\:\:\mathrm{2}{x}}&{\:\mathrm{2}{x}}&{\mathrm{1}\:+\mathrm{2}{x}}\end{vmatrix}=\:{x}^{\:\mathrm{3}} +\:\mathrm{8}{x}−\mathrm{2}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−−− \\ $$$$\:\:\: \\ $$

Answered by 1549442205PVT last updated on 17/Dec/21

△=(1+2x)^3 +4x^2 (1+2x)+8x^3 −4x^2 (1+2x)−4x^2 (1+2x)−2x(1+2x)^2   =1+6x+12x^2 +8x^3 +8x^3 −4x^2 (1+2x)  −2x−8x^2 −8x^3 =4x+1  Hence,above result is false  we can using equivalent transform of  two determinant follow as:   determinant (((1+2x   2x       2x)),((2x     1+2x     2x)),((2x     1+2x    1+2x)))∼ determinant (((1+2x   2x      2x)),((2x   1+2x     2x)),((0          0         1)))  ∼ determinant (((1+2x    2x    2x)),((−1       1        0)),((0           0          1))).Now we calculate the value of last determinant:  △=(1+2x).1.1+2x.0.0+0.(−1).2x  −0.1.2x−(−1).2x.1−0.0.(1+2x)  =2x+1+2x=4x+1

$$\bigtriangleup=\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{x}\right)+\mathrm{8}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{x}\right)−\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{x}\right)−\mathrm{2}{x}\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}+\mathrm{6}{x}+\mathrm{12}{x}^{\mathrm{2}} +\mathrm{8}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}{x}\right) \\ $$$$−\mathrm{2}{x}−\mathrm{8}{x}^{\mathrm{2}} −\mathrm{8}{x}^{\mathrm{3}} =\mathrm{4}{x}+\mathrm{1} \\ $$$${Hence},{above}\:{result}\:{is}\:{false} \\ $$$${we}\:{can}\:{using}\:{equivalent}\:{transform}\:{of} \\ $$$${two}\:{determinant}\:{follow}\:{as}: \\ $$$$\begin{vmatrix}{\mathrm{1}+\mathrm{2}{x}\:\:\:\mathrm{2}{x}\:\:\:\:\:\:\:\mathrm{2}{x}}\\{\mathrm{2}{x}\:\:\:\:\:\mathrm{1}+\mathrm{2}{x}\:\:\:\:\:\mathrm{2}{x}}\\{\mathrm{2}{x}\:\:\:\:\:\mathrm{1}+\mathrm{2}{x}\:\:\:\:\mathrm{1}+\mathrm{2}{x}}\end{vmatrix}\sim\begin{vmatrix}{\mathrm{1}+\mathrm{2}{x}\:\:\:\mathrm{2}{x}\:\:\:\:\:\:\mathrm{2}{x}}\\{\mathrm{2}{x}\:\:\:\mathrm{1}+\mathrm{2}{x}\:\:\:\:\:\mathrm{2}{x}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{vmatrix} \\ $$$$\sim\begin{vmatrix}{\mathrm{1}+\mathrm{2}{x}\:\:\:\:\mathrm{2}{x}\:\:\:\:\mathrm{2}{x}}\\{−\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{vmatrix}.{Now}\:{we}\:{calculate}\:{the}\:{value}\:{of}\:{last}\:{determinant}: \\ $$$$\bigtriangleup=\left(\mathrm{1}+\mathrm{2}{x}\right).\mathrm{1}.\mathrm{1}+\mathrm{2}{x}.\mathrm{0}.\mathrm{0}+\mathrm{0}.\left(−\mathrm{1}\right).\mathrm{2}{x} \\ $$$$−\mathrm{0}.\mathrm{1}.\mathrm{2}{x}−\left(−\mathrm{1}\right).\mathrm{2}{x}.\mathrm{1}−\mathrm{0}.\mathrm{0}.\left(\mathrm{1}+\mathrm{2}{x}\right) \\ $$$$=\mathrm{2}{x}+\mathrm{1}+\mathrm{2}{x}=\mathrm{4}{x}+\mathrm{1} \\ $$$$ \\ $$

Answered by MJS_new last updated on 17/Dec/21

⇒ 4x+1=x^3 +8x−2  ⇔ x^3 +4x−3=0  x=(((3/2)+((√(1497))/(18))))^(1/3) −((−(3/2)+((√(1497))/(28))))^(1/3) ≈.673593058  ⇒ true because indeed ∃x∈R

$$\Rightarrow\:\mathrm{4}{x}+\mathrm{1}={x}^{\mathrm{3}} +\mathrm{8}{x}−\mathrm{2} \\ $$$$\Leftrightarrow\:{x}^{\mathrm{3}} +\mathrm{4}{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{1497}}}{\mathrm{18}}}−\sqrt[{\mathrm{3}}]{−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{1497}}}{\mathrm{28}}}\approx.\mathrm{673593058} \\ $$$$\Rightarrow\:\mathrm{true}\:\mathrm{because}\:\mathrm{indeed}\:\exists{x}\in\mathbb{R} \\ $$

Commented by 1549442205PVT last updated on 18/Dec/21

Thank you Sir MjsNew,i didn′t understand the question  and give out a wrong conclusion

$${Thank}\:{you}\:{Sir}\:{MjsNew},{i}\:{didn}'{t}\:{understand}\:{the}\:{question} \\ $$$${and}\:{give}\:{out}\:{a}\:{wrong}\:{conclusion} \\ $$

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