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Question Number 21733 by Joel577 last updated on 02/Oct/17

$$\mathrm{If}p\mathrm{is}\mathrm{one}\mathrm{of}\mathrm{roots}\mathrm{from}{x}^2-2x+6=0$$$$\mathrm{then}{p}^4+16p\mathrm{is}\mathrm{equal}\mathrm{to}...$$

Answered by mrW1 last updated on 02/Oct/17

$${\mathrm{p}}^2-2\mathrm{p}+6=0$$$${\mathrm{p}}^2=2\mathrm{p}-6$$$${\mathrm{p}}^4={4\mathrm{p}}^2-24\mathrm{p}+36$$$${\mathrm{p}}^4+16\mathrm{p}={4\mathrm{p}}^2-24\mathrm{p}+36+16\mathrm{p}$$$$={4\mathrm{p}}^2-8\mathrm{p}+36$$$$=4({\mathrm{p}}^2-2\mathrm{p}+6)+12$$$$=4\times 0+12$$$$=12$$

Commented by Joel577 last updated on 02/Oct/17

$$thankyouverymuch$$

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