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Question Number 114347 by mohammad17 last updated on 18/Sep/20
if (x^2 −(h+1)x+6i−6=0) then find the value of h if you know that one of the roots of the equation is three times the square of the other root (h∈ complex number)
$$\:{if}\:\left({x}^{\mathrm{2}} −\left({h}+\mathrm{1}\right){x}+\mathrm{6}{i}−\mathrm{6}=\mathrm{0}\right)\:{then}\:{find}\:{the}\:{value}\:{of}\:{h} \\ $$$${if}\:{you}\:{know}\:{that}\:{one}\:{of}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:{is}\:{three}\: \\ $$$${times}\:{the}\:{square}\:{of}\:{the}\:{other}\:{root}\: \\ $$$$\left({h}\in\:{complex}\:{number}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by Olaf last updated on 18/Sep/20
Two roots z_1 and z_2 with z_1 = 3z_2 ^2 z_1 z_2 = (c/a) = ((6i−6)/1) = −3(√2)e^(−i(π/4)) = 3(√2)e^(i((3π)/4)) 3z_2 ^3 = 3(√2)e^(i((3π)/4)) z_2 ^3 = (√2)e^(i((3π)/4)) ⇒ z_2 = 2^(1/6) e^(i(π/4)) or 2^(1/6) e^(i((11π)/(12))) or 2^(1/6) e^(i((19π)/(12))) and z_1 = 3.2^(1/6) e^(i(π/2)) or 3.2^(1/6) e^(i((11)/6)) or 3.2^(1/6) e^(i((19π)/6)) z_1 +z_2 = −(b/a) = h+1 ⇒ h = z_1 +z_2 −1 h = 2^(1/6) (3e^(i(π/2)) +e^(i(π/4)) −1) = 2^(1/6) ((1/( (√2)))−1+i((1/( (√2)))+3)) h = 2^(1/6) (3e^(i((11π)/6)) +e^(i((11π)/(12))) −1) h = 2^(1/6) (3e^(i((19π)/6)) +e^(i((19π)/(12))) −1)
$$\mathrm{Two}\:\mathrm{roots}\:{z}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} \:\mathrm{with}\:{z}_{\mathrm{1}} \:=\:\mathrm{3}{z}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} {z}_{\mathrm{2}} \:=\:\frac{{c}}{{a}}\:=\:\frac{\mathrm{6}{i}−\mathrm{6}}{\mathrm{1}}\:=\:−\mathrm{3}\sqrt{\mathrm{2}}{e}^{−{i}\frac{\pi}{\mathrm{4}}} =\:\mathrm{3}\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \\ $$$$\mathrm{3}{z}_{\mathrm{2}} ^{\mathrm{3}} \:=\:\mathrm{3}\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \\ $$$${z}_{\mathrm{2}} ^{\mathrm{3}} \:=\:\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \\ $$$$\Rightarrow\:{z}_{\mathrm{2}} \:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} {e}^{{i}\frac{\pi}{\mathrm{4}}} \:\mathrm{or}\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} {e}^{{i}\frac{\mathrm{11}\pi}{\mathrm{12}}} \:\mathrm{or}\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} {e}^{{i}\frac{\mathrm{19}\pi}{\mathrm{12}}} \\ $$$$\mathrm{and}\:{z}_{\mathrm{1}} \:=\:\mathrm{3}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} {e}^{{i}\frac{\pi}{\mathrm{2}}} \:\mathrm{or}\:\mathrm{3}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} {e}^{{i}\frac{\mathrm{11}}{\mathrm{6}}} \:\mathrm{or}\:\mathrm{3}.\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} {e}^{{i}\frac{\mathrm{19}\pi}{\mathrm{6}}} \\ $$$${z}_{\mathrm{1}} +{z}_{\mathrm{2}} \:=\:−\frac{{b}}{{a}}\:=\:{h}+\mathrm{1} \\ $$$$\Rightarrow\:{h}\:=\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} −\mathrm{1} \\ $$$${h}\:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} \left(\mathrm{3}{e}^{{i}\frac{\pi}{\mathrm{2}}} +{e}^{{i}\frac{\pi}{\mathrm{4}}} −\mathrm{1}\right)\:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}+{i}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{3}\right)\right) \\ $$$${h}\:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} \left(\mathrm{3}{e}^{{i}\frac{\mathrm{11}\pi}{\mathrm{6}}} +{e}^{{i}\frac{\mathrm{11}\pi}{\mathrm{12}}} −\mathrm{1}\right) \\ $$$${h}\:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{6}}} \left(\mathrm{3}{e}^{{i}\frac{\mathrm{19}\pi}{\mathrm{6}}} +{e}^{{i}\frac{\mathrm{19}\pi}{\mathrm{12}}} −\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by MJS_new last updated on 19/Sep/20
good method but −6+6i=6(√2)e^((3iπ)/4) 3z_2 ^3 =6(√2)e^((3iπ)/4) ⇒ z_2 = { ((1+i)),((w(1+i))),((w^2 (1+i))) :} with w=−(1/2)+((√3)/2)i
$$\mathrm{good}\:\mathrm{method}\:\mathrm{but}\:−\mathrm{6}+\mathrm{6i}=\mathrm{6}\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{3i}\pi}{\mathrm{4}}} \\ $$$$\mathrm{3}{z}_{\mathrm{2}} ^{\mathrm{3}} =\mathrm{6}\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{3i}\pi}{\mathrm{4}}} \:\Rightarrow\:{z}_{\mathrm{2}} =\begin{cases}{\mathrm{1}+\mathrm{i}}\\{{w}\left(\mathrm{1}+\mathrm{i}\right)}\\{{w}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{i}\right)}\end{cases}\:\mathrm{with}\:{w}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$
Commented by mohammad17 last updated on 19/Sep/20
thank you sir but now whats the value of (h) can you solve this question in details please?
$${thank}\:{you}\:{sir}\:{but}\:{now}\:{whats}\:{the}\:{value}\:{of}\:\left({h}\right) \\ $$$${can}\:{you}\:{solve}\:{this}\:{question}\:{in}\:{details}\:{please}? \\ $$
Commented by MJS_new last updated on 19/Sep/20
z_1 =3z_2 ^2 = { ((6i)),((6w^2 i)),((6w^4 i=6wi)) :} z_1 +z_2 =h+1 ⇔ h=z_1 +z_2 −1= { ((7i)),((−((3−5(√3))/2)−((7−(√3))/2)i)),((−((3+5(√3))/2)−((7+(√3))/2)i)) :}
$${z}_{\mathrm{1}} =\mathrm{3}{z}_{\mathrm{2}} ^{\mathrm{2}} =\begin{cases}{\mathrm{6i}}\\{\mathrm{6}{w}^{\mathrm{2}} \mathrm{i}}\\{\mathrm{6}{w}^{\mathrm{4}} \mathrm{i}=\mathrm{6}{w}\mathrm{i}}\end{cases} \\ $$$${z}_{\mathrm{1}} +{z}_{\mathrm{2}} ={h}+\mathrm{1}\:\Leftrightarrow\:{h}={z}_{\mathrm{1}} +{z}_{\mathrm{2}} −\mathrm{1}=\begin{cases}{\mathrm{7i}}\\{−\frac{\mathrm{3}−\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{7}−\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}}\\{−\frac{\mathrm{3}+\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{7}+\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}}\end{cases} \\ $$

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