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Question Number 224435 by Hafiz00351 last updated on 11/Sep/25
Commented by TonyCWX last updated on 11/Sep/25
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Answered by TonyCWX last updated on 11/Sep/25
i. a^2 +4b^2 =a^2 −2^2 i^2 b^2 =(a+2ib)(a−2ib) ii. 9a^2 +16b^2 =3^2 a^2 −4^2 i^2 b^2 =(3a+4ib)(3a−4ib) iii. 3x^2 +3y^2 =((√3))^2 x^2 −((√3))^2 i^2 y^2 =((√3)x+(√3)iy)((√3)x−(√3)iy) iv. 144x^2 +225y^2 =12^2 x^2 −15^2 i^2 y^2 =(12x+15iy)(12x−15iy) v. z^2 −2iz−1 =(z−i)^2 vi. z^2 +6z+13 Let z^2 +6z+13=0 Then z=((−6±(√(36−4(13))))/2)=((−6±(√(−16)))/2)=((−6±4i)/2)=−3±2i So z_1 =−3+2i and z_2 −3−2i Hence z^2 +6z+13=(z−(−3+2i))(z−(−3−2i))=(z+(3−2i))(z+(3+2i)) vii. z^2 +4z+5 Let z^2 +4z+5=0 Then z=((−4±(√(16−4(5))))/2)=((−4±(√(−4)))/2)=((−4±2i)/2)=−2±i So z_1 =−2+i and z_2 =−2−i Hence z^2 +4z+5=(z−(−2+i))(z−(−2−i))=(z+(2−i))(z+(2+i)) viii. 2z^2 −22z+65 Let 2z^2 −22z+65=0 Then z=((22±(√((−22)^2 −4(2)(65))))/4)=((22±(√(−36)))/4)=((22±6i)/4)=((11±3i)/2) So z_1 =((11+3i)/2) and z_2 =((11−3i)/2) Hence 2z^2 −22z+65 = (z−((11+3i)/2))(z−((11−3i)/2))
$${i}. \\ $$$${a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} {i}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$=\left({a}+\mathrm{2}{ib}\right)\left({a}−\mathrm{2}{ib}\right) \\ $$$$ \\ $$$${ii}. \\ $$$$\mathrm{9}{a}^{\mathrm{2}} +\mathrm{16}{b}^{\mathrm{2}} \\ $$$$=\mathrm{3}^{\mathrm{2}} {a}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} {i}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$=\left(\mathrm{3}{a}+\mathrm{4}{ib}\right)\left(\mathrm{3}{a}−\mathrm{4}{ib}\right) \\ $$$$ \\ $$$${iii}. \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} \\ $$$$=\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} {i}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$=\left(\sqrt{\mathrm{3}}{x}+\sqrt{\mathrm{3}}{iy}\right)\left(\sqrt{\mathrm{3}}{x}−\sqrt{\mathrm{3}}{iy}\right) \\ $$$$ \\ $$$${iv}. \\ $$$$\mathrm{144}{x}^{\mathrm{2}} +\mathrm{225}{y}^{\mathrm{2}} \\ $$$$=\mathrm{12}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} {i}^{\mathrm{2}} {y}^{\mathrm{2}} \\ $$$$=\left(\mathrm{12}{x}+\mathrm{15}{iy}\right)\left(\mathrm{12}{x}−\mathrm{15}{iy}\right) \\ $$$$ \\ $$$${v}. \\ $$$${z}^{\mathrm{2}} −\mathrm{2}{iz}−\mathrm{1} \\ $$$$=\left({z}−{i}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${vi}. \\ $$$${z}^{\mathrm{2}} +\mathrm{6}{z}+\mathrm{13} \\ $$$${Let}\:{z}^{\mathrm{2}} +\mathrm{6}{z}+\mathrm{13}=\mathrm{0} \\ $$$${Then}\:{z}=\frac{−\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{4}\left(\mathrm{13}\right)}}{\mathrm{2}}=\frac{−\mathrm{6}\pm\sqrt{−\mathrm{16}}}{\mathrm{2}}=\frac{−\mathrm{6}\pm\mathrm{4}{i}}{\mathrm{2}}=−\mathrm{3}\pm\mathrm{2}{i} \\ $$$${So}\:{z}_{\mathrm{1}} =−\mathrm{3}+\mathrm{2}{i}\:{and}\:{z}_{\mathrm{2}} −\mathrm{3}−\mathrm{2}{i} \\ $$$${Hence}\:{z}^{\mathrm{2}} +\mathrm{6}{z}+\mathrm{13}=\left({z}−\left(−\mathrm{3}+\mathrm{2}{i}\right)\right)\left({z}−\left(−\mathrm{3}−\mathrm{2}{i}\right)\right)=\left({z}+\left(\mathrm{3}−\mathrm{2}{i}\right)\right)\left({z}+\left(\mathrm{3}+\mathrm{2}{i}\right)\right) \\ $$$$ \\ $$$${vii}. \\ $$$${z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{5} \\ $$$${Let}\:{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{5}=\mathrm{0} \\ $$$${Then}\:{z}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}\left(\mathrm{5}\right)}}{\mathrm{2}}=\frac{−\mathrm{4}\pm\sqrt{−\mathrm{4}}}{\mathrm{2}}=\frac{−\mathrm{4}\pm\mathrm{2}{i}}{\mathrm{2}}=−\mathrm{2}\pm{i} \\ $$$${So}\:{z}_{\mathrm{1}} =−\mathrm{2}+{i}\:{and}\:{z}_{\mathrm{2}} =−\mathrm{2}−{i} \\ $$$${Hence}\:{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{5}=\left({z}−\left(−\mathrm{2}+{i}\right)\right)\left({z}−\left(−\mathrm{2}−{i}\right)\right)=\left({z}+\left(\mathrm{2}−{i}\right)\right)\left({z}+\left(\mathrm{2}+{i}\right)\right) \\ $$$$ \\ $$$${viii}. \\ $$$$\mathrm{2}{z}^{\mathrm{2}} −\mathrm{22}{z}+\mathrm{65} \\ $$$${Let}\:\mathrm{2}{z}^{\mathrm{2}} −\mathrm{22}{z}+\mathrm{65}=\mathrm{0} \\ $$$${Then}\:{z}=\frac{\mathrm{22}\pm\sqrt{\left(−\mathrm{22}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}\right)\left(\mathrm{65}\right)}}{\mathrm{4}}=\frac{\mathrm{22}\pm\sqrt{−\mathrm{36}}}{\mathrm{4}}=\frac{\mathrm{22}\pm\mathrm{6}{i}}{\mathrm{4}}=\frac{\mathrm{11}\pm\mathrm{3}{i}}{\mathrm{2}} \\ $$$${So}\:{z}_{\mathrm{1}} =\frac{\mathrm{11}+\mathrm{3}{i}}{\mathrm{2}}\:{and}\:{z}_{\mathrm{2}} =\frac{\mathrm{11}−\mathrm{3}{i}}{\mathrm{2}} \\ $$$${Hence}\:\mathrm{2}{z}^{\mathrm{2}} −\mathrm{22}{z}+\mathrm{65}\:=\:\left({z}−\frac{\mathrm{11}+\mathrm{3}{i}}{\mathrm{2}}\right)\left({z}−\frac{\mathrm{11}−\mathrm{3}{i}}{\mathrm{2}}\right) \\ $$

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