Question Number 201491 by tri26112004 last updated on 07/Dec/23
$$\mathrm{1}.\:{x}^{\mathrm{2}} −\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} }=\mathrm{10}{x} \\ $$$$\mathrm{2}.\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$$$\mathrm{3}.\:\sqrt{\mathrm{2}{x}−\frac{\mathrm{8}}{{x}}}+\mathrm{2}\sqrt{\mathrm{1}−\frac{\mathrm{2}}{{x}}}\geqslant{x} \\ $$$$\mathrm{4}.\:\sqrt{{x}^{\mathrm{2}} +{x}}+\sqrt{{x}+\mathrm{2}}\geqslant\sqrt{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$\mathrm{5}.\:\left(\sqrt{{x}+\mathrm{5}}−\sqrt{{x}−\mathrm{3}}\right)\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{15}}\right)\geqslant\mathrm{8} \\ $$$$\mathrm{6}.\:{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}=\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{7}.\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{7}=\left({x}+\mathrm{5}\right)\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Dec/23
$${For}\:\mathrm{2}.\:{see}\:{also}\:{Q}#\mathrm{200498} \\ $$
Answered by Rasheed.Sindhi last updated on 08/Dec/23
$$\mathrm{2}.\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$$$\:\:\:{Let}\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:={a}\:,\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}\:={b} \\ $$$$\:\:\:\:\:\:{a}+{b}={x}\:\wedge\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} ={x}−\mathrm{1} \\ $$$$\Rightarrow{a}−{b}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}+{b}}=\frac{{x}−\mathrm{1}}{{x}} \\ $$$$\begin{cases}{\:{a}+{b}={x}}\\{{a}−{b}=\frac{{x}−\mathrm{1}}{{x}}=\mathrm{1}−\frac{\mathrm{1}}{{x}}}\end{cases}\:\: \\ $$$$\:\Rightarrow\begin{cases}{\mathrm{2}{a}={x}+\frac{{x}−\mathrm{1}}{{x}}={x}−\frac{\mathrm{1}}{{x}}+\mathrm{1}={a}^{\mathrm{2}} +\mathrm{1}}\\{\mathrm{2}{b}={x}−\frac{{x}−\mathrm{1}}{{x}}={x}+\frac{\mathrm{1}}{{x}}−\mathrm{1}={x}−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)={x}−{b}^{\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow\begin{cases}{{a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}=\mathrm{0}\Rightarrow\left({a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{a}=\mathrm{1}\Rightarrow{b}^{\mathrm{2}} +{b}={a}=\mathrm{1}}\\{{b}^{\mathrm{2}} +\mathrm{2}{b}={x}}\end{cases} \\ $$$$\Rightarrow\sqrt{{x}−\frac{\mathrm{1}}{{x}}}\:=\mathrm{1}\Rightarrow{x}−\frac{\mathrm{1}}{{x}}=\mathrm{1}\Rightarrow{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}_{\geqslant\mathrm{0}} =\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$\:\:\:\:\: \\ $$
Answered by esmaeil last updated on 07/Dec/23
$$\mathrm{1}.\:\:{x}^{\mathrm{2}} −\mathrm{1}+{x}\sqrt[{\mathrm{3}}]{{x}−\frac{\mathrm{1}}{{x}}}=\mathrm{10}{x}\Rightarrow \\ $$$${x}−\frac{\mathrm{1}}{{x}}+\sqrt[{\mathrm{3}}]{{x}−\frac{\mathrm{1}}{{x}}}=\mathrm{10} \\ $$$${x}−\frac{\mathrm{1}}{{x}}={p}^{\mathrm{3}} \Rightarrow{p}^{\mathrm{3}} +{p}=\mathrm{10}\Rightarrow \\ $$$${p}^{\mathrm{3}} −\mathrm{8}+\left({p}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\left({p}−\mathrm{2}\right)\left({p}^{\mathrm{2}} +\mathrm{2}{p}+\mathrm{5}\right)=\mathrm{0}\Rightarrow{p}=\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{1}=\mathrm{0}\Rightarrow{x}=\mathrm{4}\pm\sqrt{\mathrm{17}} \\ $$$$ \\ $$$$ \\ $$
Answered by ajfour last updated on 07/Dec/23
$$\bigstar\:{x}^{\mathrm{2}} −\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} }=\mathrm{10}{x} \\ $$$${x}−\frac{\mathrm{1}}{{x}}={p} \\ $$$${p}+\sqrt[{\mathrm{3}}]{{p}}=\mathrm{10} \\ $$$$\left(\mathrm{10}−{p}\right)^{\mathrm{3}} ={p} \\ $$$${say}\:\:\mathrm{10}−{p}={t} \\ $$$${t}^{\mathrm{3}} +{t}=\mathrm{10} \\ $$$${t}=\mathrm{2} \\ $$$${p}=\mathrm{8}={x}−\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{4}\pm\sqrt{\mathrm{17}} \\ $$$$ \\ $$
Answered by ajfour last updated on 07/Dec/23
$$\bigstar\bigstar\:\:\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$$$\sqrt{{x}−{z}}+\sqrt{\mathrm{1}−{z}}={x} \\ $$$$\mathrm{1}−{z}=\sqrt{{x}−{z}}−\sqrt{\mathrm{1}−{z}} \\ $$$$\Rightarrow\:\:\mathrm{1}−{z}+{x}=\mathrm{2}\sqrt{{x}−{z}} \\ $$$${say}\:\:\:\sqrt{{x}−{z}}={p} \\ $$$$\mathrm{1}+{p}^{\mathrm{2}} =\mathrm{2}{p} \\ $$$${p}=\mathrm{1}\:\:\Rightarrow\:\:{x}−\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Answered by Calculusboy last updated on 08/Dec/23
![Solution: 6) square both sides (x^2 +3x+1)^2 =[(x+1)(√(x^2 +1)) ]^2 (x^2 +3x)^2 +2(x^2 +3x)+1=(x+1)^2 (x^2 +1) x^4 +6x^3 +9x^2 +2x^2 +6x+1=(x^2 +2x+1)(x^2 +1) x^4 +6x^3 +11x^2 +6x+1=x^4 +2x^3 +x^2 +x^2 +2x+1 x^4 −x^4 +6x^3 −2x^3 +11x^2 −2x^2 +6x−2x+1−1=0 4x^3 +9x^2 +4x=0 (divide through by x) 4x^2 +9x+4=0 ⇔ by using x=((−b+_− (√(b^2 −4ac)))/(2a)) a=4,b=9 and c=4 x=((−9+_− (√(9^2 −4×4×4)))/(2×4)) ⇔ x=((−9+_− (√(81−64)))/8) x=((−9+_− (√(17)))/8) x_1 =((−9+(√(17)))/8) 𝚲 x_2 =((−9−(√(17)))/8)](https://www.tinkutara.com/question/Q201542.png)
$$\boldsymbol{{Solution}}: \\ $$$$\left.\mathrm{6}\right)\:\boldsymbol{{square}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}} \\ $$$$\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} =\left[\left(\boldsymbol{{x}}+\mathrm{1}\right)\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\:\right]^{\mathrm{2}} \\ $$$$\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}\right)^{\mathrm{2}} +\mathrm{2}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}\right)+\mathrm{1}=\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} \left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{6}\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{9}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}+\mathrm{1}=\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\right)\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{6}\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{11}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}+\mathrm{1}=\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}+\mathrm{1} \\ $$$$\boldsymbol{{x}}^{\mathrm{4}} −\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{6}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{11}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{6}\boldsymbol{{x}}−\mathrm{2}\boldsymbol{{x}}+\mathrm{1}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{9}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{x}}=\mathrm{0}\:\:\:\left(\boldsymbol{{divide}}\:\boldsymbol{{through}}\:\boldsymbol{{by}}\:\boldsymbol{{x}}\right) \\ $$$$\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}\boldsymbol{{x}}+\mathrm{4}=\mathrm{0}\:\:\:\:\Leftrightarrow\:\:\:\:\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{x}}=\frac{−\boldsymbol{{b}}\underset{−} {+}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{ac}}}}{\mathrm{2}\boldsymbol{{a}}} \\ $$$$\boldsymbol{{a}}=\mathrm{4},\boldsymbol{{b}}=\mathrm{9}\:\boldsymbol{{and}}\:\boldsymbol{{c}}=\mathrm{4} \\ $$$$\boldsymbol{{x}}=\frac{−\mathrm{9}\underset{−} {+}\sqrt{\mathrm{9}^{\mathrm{2}} −\mathrm{4}×\mathrm{4}×\mathrm{4}}}{\mathrm{2}×\mathrm{4}}\:\:\:\Leftrightarrow\:\:\:\:\boldsymbol{{x}}=\frac{−\mathrm{9}\underset{−} {+}\sqrt{\mathrm{81}−\mathrm{64}}}{\mathrm{8}} \\ $$$$\boldsymbol{{x}}=\frac{−\mathrm{9}\underset{−} {+}\sqrt{\mathrm{17}}}{\mathrm{8}} \\ $$$$\boldsymbol{{x}}_{\mathrm{1}} =\frac{−\mathrm{9}+\sqrt{\mathrm{17}}}{\mathrm{8}}\:\:\boldsymbol{\Lambda}\:\:\boldsymbol{{x}}_{\mathrm{2}} =\frac{−\mathrm{9}−\sqrt{\mathrm{17}}}{\mathrm{8}} \\ $$$$ \\ $$
Answered by Calculusboy last updated on 08/Dec/23
![Solution: 7) square both sides (2x^2 +3x+7)^2 =[(x+5)(√(2x^2 +1)) ]^2 (2x^2 +3x)^2 +14(2x^2 +3x)+49=(x+5)^2 (2x^2 +1) 4x^4 +12x^3 +9x^2 +28x^2 +42x+49=(x^2 +10x+25)(2x^2 +1) 4x^4 +12x^3 +37x^2 +42x+49=2x^4 +20x^3 +50x^2 +x^2 +10x+25 4x^4 −2x^4 +12x^3 −20x^3 +37x^2 −51x^2 +42x−10x+49−25=0 2x^4 −8x^3 −14x^2 +32x+24=0 (divide through by 2) x^4 −4x^3 −7x^2 +16x+12=0 (by using try and error) let x=−2 ⇒ x+2=0 (−2)^4 −4(−2)^3 −7(−2)^2 +16(−2)+12=0 ⇔ 0=0 (x+2)[x^3 −6x^2 +5x+6]=0 x_1 =−2 x^3 −6x^2 +5x+6=0 let x=2 ⇔ x−2=0 (2)^3 −6(2)^2 +5(2)+6=0 ⇔ 0=0 (x−2)[x^2 −4x−3]=0 x−2=0 ⇒ x_2 =2 x^2 −4x−3=0 ⇔ x=((−b+_− (√(b^2 −4ac)))/(2a)) a=1,b=−4 and c=−3 x=((−(−4)+_− (√((−4)^2 −4×1×(−3))))/(2×1)) x=((4+_− (√(16+12)))/2) ⇔ x=((4+_− (√(28)))/2) ⇔ x=((4+_− 2(√7))/2)=2+_− (√7) x_3 =((4+2(√7))/2)=2+(√7) 𝚲 x_4 =((4−2(√7))/2)=2−(√7) ∴ the value for (x)=(−2, 2, 2+(√7) ,2−(√7) )](https://www.tinkutara.com/question/Q201543.png)
$$\boldsymbol{{Solution}}: \\ $$$$\left.\mathrm{7}\right)\:\boldsymbol{{square}}\:\boldsymbol{{both}}\:\boldsymbol{{sides}} \\ $$$$\left(\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}+\mathrm{7}\right)^{\mathrm{2}} =\left[\left(\boldsymbol{{x}}+\mathrm{5}\right)\sqrt{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\:\right]^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}\right)^{\mathrm{2}} +\mathrm{14}\left(\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{x}}\right)+\mathrm{49}=\left(\boldsymbol{{x}}+\mathrm{5}\right)^{\mathrm{2}} \left(\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{4}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{12}\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{9}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{28}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{42}\boldsymbol{{x}}+\mathrm{49}=\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{10}\boldsymbol{{x}}+\mathrm{25}\right)\left(\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{4}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{12}\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{37}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{42}\boldsymbol{{x}}+\mathrm{49}=\mathrm{2}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{20}\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{50}\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{10}\boldsymbol{{x}}+\mathrm{25} \\ $$$$\mathrm{4}\boldsymbol{{x}}^{\mathrm{4}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{12}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{20}\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{37}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{51}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{42}\boldsymbol{{x}}−\mathrm{10}\boldsymbol{{x}}+\mathrm{49}−\mathrm{25}=\mathrm{0} \\ $$$$\mathrm{2}\boldsymbol{{x}}^{\mathrm{4}} −\mathrm{8}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{14}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{32}\boldsymbol{{x}}+\mathrm{24}=\mathrm{0}\:\:\:\left(\boldsymbol{{divide}}\:\boldsymbol{{through}}\:\boldsymbol{{by}}\:\mathrm{2}\right) \\ $$$$\boldsymbol{{x}}^{\mathrm{4}} −\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{7}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{16}\boldsymbol{{x}}+\mathrm{12}=\mathrm{0}\:\:\:\:\:\left(\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{try}}\:\boldsymbol{{and}}\:\boldsymbol{{error}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{x}}=−\mathrm{2}\:\:\Rightarrow\:\boldsymbol{{x}}+\mathrm{2}=\mathrm{0} \\ $$$$\left(−\mathrm{2}\right)^{\mathrm{4}} −\mathrm{4}\left(−\mathrm{2}\right)^{\mathrm{3}} −\mathrm{7}\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{16}\left(−\mathrm{2}\right)+\mathrm{12}=\mathrm{0}\:\:\Leftrightarrow\:\:\mathrm{0}=\mathrm{0} \\ $$$$\left(\boldsymbol{{x}}+\mathrm{2}\right)\left[\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{6}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}+\mathrm{6}\right]=\mathrm{0} \\ $$$$\boldsymbol{{x}}_{\mathrm{1}} =−\mathrm{2} \\ $$$$\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{6}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}+\mathrm{6}=\mathrm{0} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{x}}=\mathrm{2}\:\:\Leftrightarrow\:\:\boldsymbol{{x}}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)^{\mathrm{3}} −\mathrm{6}\left(\mathrm{2}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{2}\right)+\mathrm{6}=\mathrm{0}\:\:\Leftrightarrow\:\:\mathrm{0}=\mathrm{0} \\ $$$$\left(\boldsymbol{{x}}−\mathrm{2}\right)\left[\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}−\mathrm{3}\right]=\mathrm{0} \\ $$$$\boldsymbol{{x}}−\mathrm{2}=\mathrm{0}\:\:\Rightarrow\:\boldsymbol{{x}}_{\mathrm{2}} =\mathrm{2} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}−\mathrm{3}=\mathrm{0}\:\:\Leftrightarrow\:\:\boldsymbol{{x}}=\frac{−\boldsymbol{{b}}\underset{−} {+}\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{ac}}}}{\mathrm{2}\boldsymbol{{a}}} \\ $$$$\boldsymbol{{a}}=\mathrm{1},\boldsymbol{{b}}=−\mathrm{4}\:\boldsymbol{{and}}\:\boldsymbol{{c}}=−\mathrm{3} \\ $$$$\boldsymbol{{x}}=\frac{−\left(−\mathrm{4}\right)\underset{−} {+}\sqrt{\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\left(−\mathrm{3}\right)}}{\mathrm{2}×\mathrm{1}} \\ $$$$\boldsymbol{{x}}=\frac{\mathrm{4}\underset{−} {+}\sqrt{\mathrm{16}+\mathrm{12}}}{\mathrm{2}}\:\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{x}}=\frac{\mathrm{4}\underset{−} {+}\sqrt{\mathrm{28}}}{\mathrm{2}}\:\:\Leftrightarrow\:\:\boldsymbol{{x}}=\frac{\mathrm{4}\underset{−} {+}\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{2}}=\mathrm{2}\underset{−} {+}\sqrt{\mathrm{7}} \\ $$$$\boldsymbol{{x}}_{\mathrm{3}} =\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{7}}\:\:\:\boldsymbol{\Lambda}\:\:\:\boldsymbol{{x}}_{\mathrm{4}} =\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{2}}=\mathrm{2}−\sqrt{\mathrm{7}} \\ $$$$\therefore\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{for}}\:\left(\boldsymbol{{x}}\right)=\left(−\mathrm{2},\:\mathrm{2},\:\mathrm{2}+\sqrt{\mathrm{7}}\:,\mathrm{2}−\sqrt{\mathrm{7}}\:\right) \\ $$$$ \\ $$$$ \\ $$