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Question Number 147894 by mathdanisur last updated on 24/Jul/21

$$x^2-3x+4=0\Rightarrow x^2+\frac{16}{x^2}=?$$

Answered by mindispower last updated on 24/Jul/21

$$\iff x+\frac{4}{x}=3$$$$\Rightarrow x^2+\frac{16}{x^2}+8=9$$

Commented by mathdanisur last updated on 24/Jul/21

$$ThankyouSir$$

Answered by puissant last updated on 24/Jul/21

$${\mathrm{x}}^2-3\mathrm{x}+4=0\Rightarrow {\mathrm{x}}^2+4=3\mathrm{x}$$$$\Rightarrow \mathrm{x}+\frac{4}{\mathrm{x}}=3$$$$\Rightarrow {(\mathrm{x}+\frac{4}{\mathrm{x}})}^2=9$$$$\Rightarrow {\mathrm{x}}^2+2\mathrm{x}\frac{4}{\mathrm{x}}+\frac{16}{{\mathrm{x}}^2}=9$$$$\Rightarrow {\mathrm{x}}^2+\frac{16}{{\mathrm{x}}^2}=9-8=1....$$

Commented by mathdanisur last updated on 24/Jul/21

$$ThankyouSir$$

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