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Question Number 96407 by Enyz last updated on 01/Jun/20

Commented by prakash jain last updated on 01/Jun/20

$$z^3+2z^2+(k-8\sqrt{2}i)z+(8-4k\sqrt{2}i)=0$$$$\mathrm{Let}z=x\mathrm{be}\mathrm{real}\mathrm{root}$$$$x^3+2x^2+kx-8ix\sqrt{2}+8-4k\sqrt{2}i=0\left(\mathrm{i}\right)$$$$x\mathrm{is}\mathrm{real}$$$$\mathrm{Imaginary}\mathrm{part}=0$$$$-8x\sqrt{2}+4k\sqrt{2}=0$$$$x=-\frac{k}{2}$$$$\mathrm{real}\mathrm{part}$$$$x^3+2x^2+kx+8=0$$$$\mathrm{put}x=-\frac{k}{2}$$$$-k^3+4k^2-k^2+16=0$$$$k^3-3k^2-16=0$$$$k^3-4k^2+k^2-16=0$$$$k^2(k-4)+(k-4)(k+4)=0$$$$(k^2+k+4)(k-4)=0$$$$\mathrm{only}\mathrm{real}\mathrm{value}k=4\mathrm{satisfies}.$$$$\Rightarrow k=4$$$$k=4\mathrm{in}\mathrm{original}\mathrm{equation}$$$$z^3+2z^2+(4-8\sqrt{2}i)z+(8-16\sqrt{2}i)=0$$$$z^2(z+2)+(4-8i\sqrt{2})(z+2)=0$$$$(z+2)(z^2+4-8i\sqrt{2})=0$$$$z=-2\mathrm{already}\mathrm{considered}$$$$\mathrm{other}\mathrm{two}\mathrm{roots}$$$$z^2=-4+8i\sqrt{2}$$$$z=\pm \sqrt{-4+8i\sqrt{2}}$$$$z=\pm 2\sqrt{-1+2i\sqrt{2}}$$

Commented by Enyz last updated on 01/Jun/20

Thanks a lot Mr Prakash. But when I substitute the complex roots into the orginal equation with k=4, it doesnt come to 0. Can you spot my mistake?

Commented by peter frank last updated on 01/Jun/20

$$\mathrm{thank}\mathrm{you}$$

Commented by peter frank last updated on 01/Jun/20

$$\mathrm{thank}\mathrm{you}$$

Answered by Enyz last updated on 01/Jun/20

Commented by Enyz last updated on 01/Jun/20

Could someone check whether this answer Is CORRECT?

Commented by prakash jain last updated on 01/Jun/20

$$\mathrm{Please}\mathrm{see}\mathrm{comments}.\mathrm{You}\mathrm{seem}\mathrm{to}$$$$\mathrm{have}\mathrm{utilized}\mathrm{that}\mathrm{fact}z=-2\mathrm{is}\mathrm{root}$$$$\mathrm{to}\mathrm{determine}k.$$$$\mathrm{How}\mathrm{would}\mathrm{you}\mathrm{approach}\mathrm{the}\mathrm{problem}$$$$\mathrm{if}\mathrm{you}\mathrm{did}\mathrm{not}\mathrm{know}z=-2\mathrm{is}$$$$\mathrm{a}\mathrm{real}\mathrm{root}?$$

Commented by prakash jain last updated on 01/Jun/20

$$(a+bi)\mathrm{and}\mathrm{you}\mathrm{are}\mathrm{assuming}a\mathrm{is}$$$$\mathrm{real}\mathrm{and}\mathrm{solving}\mathrm{for}\mathrm{complex}\mathrm{value}$$$$\mathrm{that}\mathrm{is}\mathrm{not}\mathrm{correct}.$$$$\mathrm{After}\mathrm{factoring}(z+2)\mathrm{just}\mathrm{use}$$$$\mathrm{quadratic}\mathrm{formula}.$$

Commented by Enyz last updated on 01/Jun/20

since the equation has one real root, the other two roots are complex conjugates. But my final answer does not fit into the original eqn with k=4

Commented by Enyz last updated on 01/Jun/20

Wondering what went wrong..

Commented by Enyz last updated on 01/Jun/20

Glad if u could help me get the other 2 roots.

Commented by prakash jain last updated on 01/Jun/20

$$\mathrm{while}\mathrm{assumin}a+ib=z$$$$a\mathrm{and}b\mathrm{both}\mathrm{are}\mathrm{real}$$$$a=2i\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{valid}\mathrm{solution}.$$$$\mathrm{Hence}\mathrm{you}\mathrm{are}\mathrm{getting}\mathrm{incorrect}\mathrm{annswer}.$$

Commented by prakash jain last updated on 01/Jun/20

Commented by prakash jain last updated on 01/Jun/20

$$\mathrm{Please}\mathrm{note}:$$$$\mathrm{A}\mathrm{quadratic}\mathrm{equation}\mathrm{has}2$$$$\mathrm{complex}\mathrm{conjugate}\mathrm{roots}.\mathrm{This}$$$$\mathrm{is}\mathrm{ONLY}\mathrm{true}\mathrm{if}\mathrm{coefficients}\mathrm{are}$$$$\mathrm{REAL}.$$$$(z+i)(z-2+3i)=0$$$$\mathrm{The}\mathrm{above}\mathrm{equation}\mathrm{has}$$$$2\mathrm{complex}\mathrm{roots}\mathrm{which}\mathrm{are}\mathrm{not}$$$$\mathrm{conjugates}.$$

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