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Question Number 174092 by behi834171 last updated on 24/Jul/22

$${\mathit{x}}^4+{\mathit{a}\mathit{x}}^2+14\mathit{x}-210=0$$$${\mathit{x}}_1.{\mathit{x}}_2=22,\mathit{a}=?$$

Answered by Rasheed.Sindhi last updated on 24/Jul/22

$$x_2=\frac{22}{x_1}$$$$x_1^4+{ax}_1^2+14x_1-210={\left(\frac{22}{x_1}\right)}^4+a{\left(\frac{22}{x_1}\right)}^2+14\left(\frac{22}{x_1}\right)-210=0$$$$x_1^4+{ax}_1^2+14x_1-210=\frac{{22}^4}{x_1^4}+a\left(\frac{{22}^2}{x_1^2}\right)+14\left(\frac{22}{x_1}\right)-210=0$$$$x_1^8+{ax}_1^6+14x_1^5-210x_1^4={22}^4+{22}^2{ax}_1^2+14\left(22\right)x_1^3-210x_1^4=0$$$$x_1^8+{ax}_1^6+14x_1^5={22}^4+{22}^2{ax}_1^2+14\left(22\right)x_1^3=0$$$$$$$$Continue$$

Commented by behi834171 last updated on 24/Jul/22

$$waitingforfinalresualts.....dearmaster$$

Commented by MJS_new last updated on 24/Jul/22

$$\mathrm{the}\mathrm{key}\mathrm{is}$$$$x^4+0x^3+{px}^2+qx+r$$$$\mathrm{always}\mathrm{has}\mathrm{two}\mathrm{square}\mathrm{factors}$$$$(x^2-\alpha x-\beta )(x^2+\alpha x-\gamma )$$

Commented by Rasheed.Sindhi last updated on 25/Jul/22

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathrm{for}\mathrm{real}\mathrm{help}\mathbf{sir}!$$

Answered by MJS_new last updated on 24/Jul/22

$$\mathrm{let}{x}_1=p\wedge x_2=\frac{22}{p}$$$$\Rightarrow$$$$x^4+{ax}^2+14x-210=(x^2-\frac{p^2+22}{p}x+22)(x^2+\frac{p^2+22}{p}x-\frac{105}{11})$$$$\mathrm{since}p\ne 0\mathrm{we}\mathrm{get}\mathrm{by}\mathrm{matching}\mathrm{constants}$$$$\{\begin{array}{l}{p}^2-\frac{154}{347}p+22=0\\ a=-\frac{11p^4+347p^2+5324}{11p^2}\end{array}$$$$\Rightarrow$$$$p=\frac{77}{347}\pm \frac{\sqrt{2643069}}{347}\mathrm{i}$$$$a=\frac{16235157}{1324499}$$

Commented by behi834171 last updated on 24/Jul/22

$$thankyouverymuchdearmaster.$$

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