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Question Number 95262 by 6532 last updated on 24/May/20

$${3\mathrm{x}}^2+{5\mathrm{x}}^4-7$$$$\mathrm{plz}\mathrm{help}\mathrm{to}\mathrm{solve}\mathrm{this}\mathrm{equation}$$

Answered by i jagooll last updated on 24/May/20

$$\mathrm{do}\mathrm{you}\mathrm{meant}{3\mathrm{x}}^2+{5\mathrm{x}}^4-7=0?$$

Commented by 6532 last updated on 24/May/20

$$\mathrm{ohh}\mathrm{yes}\mathrm{sorry}\mathrm{i}\mathrm{made}\mathrm{a}\mathrm{mistake}$$

Answered by bobhans last updated on 24/May/20

$$\mathrm{set}{\mathrm{x}}^2=\mathrm{t},\mathrm{t}⩾0\mathrm{for}\mathrm{x}\in \mathbb{R}$$$${5\mathrm{t}}^2+3\mathrm{t}-7=0$$$$\mathrm{t}=\frac{-3\pm \sqrt{9+140}}{10}=\frac{-3\pm \sqrt{149}}{10}$$$$\mathrm{may}\mathrm{be}\mathrm{we}\mathrm{use}\mathrm{a}\mathrm{calculator}$$

Answered by MJS last updated on 24/May/20

$$5x^4+3x^2-7=0$$$$x^4+\frac{3}{5}{x}^2-\frac{7}{5}=0$$$${\left(x^2\right)}^2+\frac{3}{5}\left(x^2\right)-\frac{7}{5}=0$$$$x^2=-\frac{3\pm \sqrt{149}}{10}$$$$x=\pm \frac{\sqrt{-30+10\sqrt{149}}}{10}\vee x=\pm \frac{\sqrt{30+10\sqrt{149}}}{10}\mathrm{i}$$

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