Question-195433

Question Number 195433 by sonukgindia last updated on 02/Aug/23

Answered by gatocomcirrose last updated on 02/Aug/23

2 determinant ((2,(−2),1),(x,2,(−1)),(1,(−1),(−2)))−3 determinant ((x,(−2),1),(2,2,(−1)),(x,(−1),(−2)))+ +i determinant ((x,2,1),(2,x,(−1)),(x,1,(−2)))+ determinant ((x,2,(−2)),(2,x,2),(x,1,(−1)))=0 ⇒2[−10−5x]−3[−5x−10]+i[−3x^2 −x+10]+[x^2 +2x]=0 ⇒x^2 (1−3i)+x(7−i)+10+10i=0 x=((i−7±(√(−110+66i)))/(2−6i))

$$\mathrm{2}\begin{vmatrix}{\mathrm{2}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{x}}&{\mathrm{2}}&{−\mathrm{1}}\\{\mathrm{1}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}−\mathrm{3}\begin{vmatrix}{\mathrm{x}}&{−\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{2}}&{−\mathrm{1}}\\{\mathrm{x}}&{−\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}+ \\ $$$$+\mathrm{i}\begin{vmatrix}{\mathrm{x}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{x}}&{−\mathrm{1}}\\{\mathrm{x}}&{\mathrm{1}}&{−\mathrm{2}}\end{vmatrix}+\begin{vmatrix}{\mathrm{x}}&{\mathrm{2}}&{−\mathrm{2}}\\{\mathrm{2}}&{\mathrm{x}}&{\mathrm{2}}\\{\mathrm{x}}&{\mathrm{1}}&{−\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}\left[−\mathrm{10}−\mathrm{5x}\right]−\mathrm{3}\left[−\mathrm{5x}−\mathrm{10}\right]+\mathrm{i}\left[−\mathrm{3x}^{\mathrm{2}} −\mathrm{x}+\mathrm{10}\right]+\left[\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{3i}\right)+\mathrm{x}\left(\mathrm{7}−\mathrm{i}\right)+\mathrm{10}+\mathrm{10i}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{i}−\mathrm{7}\pm\sqrt{−\mathrm{110}+\mathrm{66i}}}{\mathrm{2}−\mathrm{6i}} \\ $$

Commented by Frix last updated on 02/Aug/23

Who gives a “like” to a wrong answer? (1−3i)x^2 +(7−i)x+10(1+i)=0 x^2 +(1+2i)x−2(1−2i)=0 (x+2)(x−(1−2i))=0 x_1 =−2 x_2 =1−2i

$$\mathrm{Who}\:\mathrm{gives}\:\mathrm{a}\:“\mathrm{like}''\:\mathrm{to}\:\mathrm{a}\:\mathrm{wrong}\:\mathrm{answer}? \\ $$$$\left(\mathrm{1}−\mathrm{3i}\right){x}^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{i}\right){x}+\mathrm{10}\left(\mathrm{1}+\mathrm{i}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{2i}\right){x}−\mathrm{2}\left(\mathrm{1}−\mathrm{2i}\right)=\mathrm{0} \\ $$$$\left({x}+\mathrm{2}\right)\left({x}−\left(\mathrm{1}−\mathrm{2i}\right)\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\mathrm{2} \\ $$$${x}_{\mathrm{2}} =\mathrm{1}−\mathrm{2i} \\ $$

Commented by Frix last updated on 02/Aug/23

Only one mistake (typo?) x=((−7+i±(√(−112+66i)))/(2−6i))=((−7+i±(3+11i))/(2−6i))= = { ((((−10−10i)/(2−6i))=1−2i)),((((−4+12i)/(2−6i))=−2)) :}

$$\mathrm{Only}\:\mathrm{one}\:\mathrm{mistake}\:\left(\mathrm{typo}?\right) \\ $$$${x}=\frac{−\mathrm{7}+\mathrm{i}\pm\sqrt{−\mathrm{112}+\mathrm{66i}}}{\mathrm{2}−\mathrm{6i}}=\frac{−\mathrm{7}+\mathrm{i}\pm\left(\mathrm{3}+\mathrm{11i}\right)}{\mathrm{2}−\mathrm{6i}}= \\ $$$$=\begin{cases}{\frac{−\mathrm{10}−\mathrm{10i}}{\mathrm{2}−\mathrm{6i}}=\mathrm{1}−\mathrm{2i}}\\{\frac{−\mathrm{4}+\mathrm{12i}}{\mathrm{2}−\mathrm{6i}}=−\mathrm{2}}\end{cases} \\ $$

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