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Question Number 190169 by sonukgindia last updated on 29/Mar/23

Commented by Frix last updated on 29/Mar/23

$$\mathrm{Folks}2\mathrm{questions}:$$$$\left(1\right)\mathrm{How}\mathrm{many}\mathrm{solutions}\mathrm{does}\mathrm{a}\mathrm{polynome}$$$$\mathrm{of}\mathrm{degree}n\mathrm{have}?$$$$\left(2\right)\mathrm{Do}\mathrm{you}\mathrm{ever}\mathrm{test}\mathrm{your}\mathrm{solutions}?$$

Commented by Frix last updated on 29/Mar/23

$${ax}^2+bx+c=0;a,b,c,x\in \mathbb{C}\Rightarrow$$$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$$$\mathrm{Only}\mathrm{MrW}\mathrm{seems}\mathrm{to}\mathrm{know}\mathrm{basic}\mathrm{formulas}$$

Commented by Frix last updated on 29/Mar/23

$$8x^2\mathrm{i}-14x=\frac{15}{i}$$$$-8x^2-14\mathrm{i}x-15=0$$$$8x^2+14\mathrm{i}x+15=0$$$$x=\frac{-14\mathrm{i}\pm \sqrt{{\left(14\mathrm{i}\right)}^2-4\times 8\times 15}}{2\times 8}$$$$x=\frac{-14\mathrm{i}\pm \sqrt{-196-480}}{16}$$$$x=\frac{-14\mathrm{i}\pm 26\mathrm{i}}{16}=\{\begin{array}{l}\frac{3}{4}\mathrm{i}\\ -\frac{5}{2}\mathrm{i}\end{array}$$

Answered by Rasheed.Sindhi last updated on 29/Mar/23

$$Letx=a+bi$$$$8i{(a+bi)}^2-14(a+bi)=\frac{15}{i}\cdot \frac{-i}{-i}$$$$8i(a^2-b^2+2abi)-14a-14bi=-15i$$$$8i(a^2-b^2)+16{abi}^2-14a-14bi=-15i$$$$8i(a^2-b^2)-14bi-16ab-14a=-15i$$$$(8a^2-8b^2-14b)i-16ab-14a=0-15i$$$$-16ab-14a=0\wedge 8a^2-8b^2-14b=-15$$$$\bullet a(-8b-7)=0\Rightarrow a=0\vee -8b-7=0$$$$\underset{―}{\bullet a=0_{}:}$$$$8a^2-8b^2-14b=-15$$$$8{\left(0\right)}^2-8b^2-14b=-15$$$$8b^2-14b+15=0$$$$b=\frac{7}{8}\pm \frac{\sqrt{71}}{8}i$$$$x=a+bi=0+(\frac{7}{8}\pm \frac{\sqrt{71}}{8}i)i$$$$=\mp \frac{\sqrt{71}}{8}+\frac{7}{8}i$$$$\underset{―}{\bullet -8b-7=0_{}}\Rightarrow b=\frac{7}{8}:$$$$8a^2-8b^2-14b=-15$$$$8a^2-8{\left(\frac{7}{8}\right)}^2-14\left(\frac{7}{8}\right)=-15$$$$8a^2-\frac{49}{8}-\frac{98}{8}=-15$$$$8a^2=-15+\frac{147}{8}=\frac{27}{8}$$$$a=\pm \frac{3\sqrt{3}}{8}$$$$x=a+bi=\pm \frac{3\sqrt{3}}{8}+\frac{7}{8}i$$

Commented by JDamian last updated on 29/Mar/23

there is a -14b that it should be +14b

Commented by mehdee42 last updated on 29/Mar/23

$$Letx=a+bi$$$$8i{(a+bi)}^2-14(a+bi)=\frac{15}{i}\cdot \frac{-i}{-i}$$$$8i(a^2-b^2+2abi)-14a-14bi=-15i$$$$8i(a^2-b^2)+16{abi}^2-14a-14bi=-15i$$$$8i(a^2-b^2)-14bi-16ab-14a=-15i$$$$(8a^2-8b^2-14b)i-16ab-14a=0-15i$$$$-16ab-14a=0\wedge 8a^2-8b^2-14b=-15$$$$\bullet a(-8b-7)=0\Rightarrow a=0\vee -8b-7=0$$$$\underset{―}{\bullet a=0_{}:}$$$$8a^2-8b^2-14b=-15$$$$8{\left(0\right)}^2-8b^2-14b=-15$$$$\nRightarrow 8b^2-14b+15=0$$$$(-8b^2-14b+15=0)\Rightarrow b=\frac{3}{4}\vee -\frac{5}{2}$$$$\Rightarrow x=\frac{3}{4}i\vee -\frac{5}{2}i$$$$b=\frac{7}{8}\pm \frac{\sqrt{71}}{8}i$$$$x=a+bi=0+(\frac{7}{8}\pm \frac{\sqrt{71}}{8}i)i$$$$=\mp \frac{\sqrt{71}}{8}+\frac{7}{8}i$$$$\underset{―}{\bullet -8b-7=0_{}}\Rightarrow b=\nRightarrow \frac{7}{8}:b=-\frac{7}{8}\Rightarrow a^2=-\frac{71}{64}thereisnoanswer$$$$8a^2-8b^2-14b=-15$$$$8a^2-8{\left(\frac{7}{8}\right)}^2-14\left(\frac{7}{8}\right)=-15$$$$8a^2-\frac{49}{8}-\frac{98}{8}=-15$$$$8a^2=-15+\frac{147}{8}=\frac{27}{8}$$$$a=\pm \frac{3\sqrt{3}}{8}$$$$x=a+bi=\pm \frac{3\sqrt{3}}{8}+\frac{7}{8}i$$

Commented by Frix last updated on 29/Mar/23

$$\mathrm{So}\mathrm{you}\mathrm{get}4\mathrm{solutions}?!$$$$\mathrm{You}\mathrm{seriously}\mathrm{claim}\mathrm{that}:$$$${ax}^2+bx+c=$$$$=a(x-x_1)(x-x_2)(x-x_3)(x-x_4)$$$$\iff$$$${ax}^2+bx+c=$$$$=a(x^4+{Ax}^3+{Bx}^2+Cx+D)$$$$\mathrm{A}\mathrm{miracle}!$$

Commented by Rasheed.Sindhi last updated on 29/Mar/23

$$Nomiraclesir:)$$$$Somesolutionsareinvalid.Ishould$$$$havetested.Anywaymymistakesir!$$

Commented by Rasheed.Sindhi last updated on 29/Mar/23

$$@mahdee$$$$Thankssirtopointoutmymistake!$$

Commented by MJS_new last updated on 29/Mar/23

$$\mathrm{Sir}\mathrm{Rasheed},\mathrm{look}\mathrm{at}\mathrm{my}\mathrm{solution}.\mathrm{The}\mathrm{error}$$$$\mathrm{is}\mathrm{if}\mathrm{you}\mathrm{let}x=a+b\mathrm{i}\mathrm{then}a,b\mathrm{must}\mathrm{be}\mathrm{real}.$$

Commented by Rasheed.Sindhi last updated on 29/Mar/23

$$e^{x}cellentpoint\mathit{s}\mathit{i}\mathit{r}!\mathcal{T}han\mathcal{X}!$$

Commented by mehdee42 last updated on 29/Mar/23

$$IThankyoutoo.yoursolutioniscertainly$$$$verybeautiful.$$

Commented by Rasheed.Sindhi last updated on 30/Mar/23

��

Answered by mr W last updated on 29/Mar/23

$$x=\frac{14\pm \sqrt{{14}^2+4\times 8i\times \frac{15}{i}}}{2\times 8i}$$$$=\frac{7\pm 13}{8i}$$$$=-\left(\frac{7\pm 13}{8}\right)i=-\frac{5i}{2}or\frac{3i}{4}$$

Commented by Rasheed.Sindhi last updated on 31/Mar/23

$$\mathbb{N}\mathbf{i}\subset \in \mathbf{Sir}!$$

Answered by MJS_new last updated on 29/Mar/23

$$\mathrm{if}\mathrm{we}\mathrm{let}x=a+b\mathrm{i};a,b\in \mathbb{R}\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{system}$$$$\{\begin{array}{l}2a(8b+7)=0\\ 8a^2-8b^2-14b+15=0\end{array}$$$$\left(1\right)\Rightarrow a=0\vee b=-\frac{7}{8}$$$$1.a=0$$$$\Rightarrow \left(2\right)(b+\frac{5}{2})(b-\frac{3}{4})=0$$$$\Rightarrow b=-\frac{5}{2}\vee b=\frac{3}{4}$$$$\mathrm{both}\mathrm{are}\mathrm{valid}\mathrm{because}\mathrm{they}\mathrm{fit}\left(1\right)\mathrm{and}\left(2\right)$$$$\mathrm{when}a=0$$$$2.b=-\frac{7}{8}$$$$\Rightarrow \left(2\right)a^2=-\frac{169}{64}$$$$\Rightarrow a\notin \mathbb{R}\iff \mathrm{no}\mathrm{solution}$$$$\Rightarrow a=0\wedge (b=-\frac{5}{2}\vee b=\frac{3}{4})$$$$\Rightarrow x=-\frac{5}{2}\mathrm{i}\vee x=\frac{3}{4}\mathrm{i}$$

Commented by Rasheed.Sindhi last updated on 29/Mar/23

$$\cap \mathbf{i}\subset \in !$$$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbf{sir}!$$

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