Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 151468 by peter frank last updated on 21/Aug/21

z^2 −7z+16=i(z−11)

$$\mathrm{z}^{\mathrm{2}} −\mathrm{7z}+\mathrm{16}=\mathrm{i}\left(\mathrm{z}−\mathrm{11}\right) \\ $$$$ \\ $$

Answered by peter frank last updated on 21/Aug/21

z^2 −7z−iz+16+11i=0  z^2 −(7+i)z+16+11i=0  z=(((7+i)±(√((7+i)^2 −4(1)(16+11i))))/2)  ......

$$\mathrm{z}^{\mathrm{2}} −\mathrm{7z}−\mathrm{iz}+\mathrm{16}+\mathrm{11i}=\mathrm{0} \\ $$$$\mathrm{z}^{\mathrm{2}} −\left(\mathrm{7}+\mathrm{i}\right)\mathrm{z}+\mathrm{16}+\mathrm{11i}=\mathrm{0} \\ $$$$\mathrm{z}=\frac{\left(\mathrm{7}+\mathrm{i}\right)\pm\sqrt{\left(\mathrm{7}+\mathrm{i}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{16}+\mathrm{11i}\right)}}{\mathrm{2}} \\ $$$$...... \\ $$

Commented by MJS_new last updated on 22/Aug/21

you get the same  ...  z=(((7+i)±(√(−16−30i)))/2)  (√(−16−30i))=3−5i  ⇒  z=2+3i∨z=5−2i

$$\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same} \\ $$$$... \\ $$$${z}=\frac{\left(\mathrm{7}+\mathrm{i}\right)\pm\sqrt{−\mathrm{16}−\mathrm{30i}}}{\mathrm{2}} \\ $$$$\sqrt{−\mathrm{16}−\mathrm{30i}}=\mathrm{3}−\mathrm{5i} \\ $$$$\Rightarrow \\ $$$${z}=\mathrm{2}+\mathrm{3i}\vee{z}=\mathrm{5}−\mathrm{2i} \\ $$

Answered by MJS_new last updated on 21/Aug/21

z^2 −(7+i)z+16+11i=0  z=t+((7+i)/2)  t^2 +4+((15)/2)i=0  t=±(√(−4−((15)/2)i))=±((3/2)−(5/2)i)  z=2+3i∨z=5−2i

$${z}^{\mathrm{2}} −\left(\mathrm{7}+\mathrm{i}\right){z}+\mathrm{16}+\mathrm{11i}=\mathrm{0} \\ $$$${z}={t}+\frac{\mathrm{7}+\mathrm{i}}{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} +\mathrm{4}+\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i}=\mathrm{0} \\ $$$${t}=\pm\sqrt{−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i}}=\pm\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{i}\right) \\ $$$${z}=\mathrm{2}+\mathrm{3i}\vee{z}=\mathrm{5}−\mathrm{2i} \\ $$

Commented by peter frank last updated on 22/Aug/21

more step  sir not understood

$$\mathrm{more}\:\mathrm{step}\:\:\mathrm{sir}\:\mathrm{not}\:\mathrm{understood} \\ $$

Commented by MJS_new last updated on 22/Aug/21

x^2 +px+q=0  let x=t−(p/2)  (just another version to get the usual term]  t^2 −(p^2 /4)+q=0  t=±(√((p^2 /4)−q))  p=−(7+i)∧q=16+11i  ⇒  t=±(√(−4−((15)/2)i))  (a+bi)^2 =−4−((15)/2)i  a^2 −b^2 +2abi=−4−((15)/2)i  ⇒  a^2 −b^2 =−4∧2ab=−((15)/2) ⇒ a=±(3/2)∧b=∓(5/2)  ±(√(−4−((15)/2)i))=±((3/2)−(5/2)i)

$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}={t}−\frac{{p}}{\mathrm{2}}\:\:\left(\mathrm{just}\:\mathrm{another}\:\mathrm{version}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{usual}\:\mathrm{term}\right] \\ $$$${t}^{\mathrm{2}} −\frac{{p}^{\mathrm{2}} }{\mathrm{4}}+{q}=\mathrm{0} \\ $$$${t}=\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}} \\ $$$${p}=−\left(\mathrm{7}+\mathrm{i}\right)\wedge{q}=\mathrm{16}+\mathrm{11i} \\ $$$$\Rightarrow \\ $$$${t}=\pm\sqrt{−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i}} \\ $$$$\left({a}+{b}\mathrm{i}\right)^{\mathrm{2}} =−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}\mathrm{i}=−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =−\mathrm{4}\wedge\mathrm{2}{ab}=−\frac{\mathrm{15}}{\mathrm{2}}\:\Rightarrow\:{a}=\pm\frac{\mathrm{3}}{\mathrm{2}}\wedge{b}=\mp\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\pm\sqrt{−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i}}=\pm\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{i}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com