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Question Number 83608 by john santu last updated on 04/Mar/20

$$\sqrt[4]{17+\mathrm{x}}+\sqrt[4]{17-\mathrm{x}}=2$$$$\mathrm{find}\mathrm{x}$$

Commented by mr W last updated on 04/Mar/20

$$LHS⩾\sqrt[4]{2\times 17}>\sqrt[4]{2\times 16}>\sqrt[4]{16}=2=RHS$$$$\Rightarrow nosolutionforLHS=RHS$$

Commented by jagoll last updated on 04/Mar/20

$$\mathrm{yes}.\mathrm{i}\mathrm{got}\mathrm{x}=\varnothing$$

Commented by john santu last updated on 04/Mar/20

$$\mathrm{okay}$$

Commented by MJS last updated on 04/Mar/20

$$x=a+b\mathrm{i}$$$$\mathrm{scetch}\mathrm{it};\mathrm{let}b>0[\mathrm{similar}\mathrm{for}b<0]$$$$z_1=17+a+b\mathrm{i}\mathrm{lies}\mathrm{above}\mathrm{the}\mathrm{real}\mathrm{axis}$$$$z_2=17-a-b\mathrm{i}\mathrm{lies}\mathrm{under}\mathrm{the}\mathrm{real}\mathrm{axis}$$$$\mathrm{the}\mathrm{angles}\mathrm{of}{z}_1\mathrm{and}{z}_2\mathrm{are}\mathrm{only}\mathrm{of}\pm \mathrm{the}\mathrm{same}$$$$\mathrm{values}\mathrm{if}a=0.\mathrm{but}\mathrm{then}\mid z_1^{\frac{1}{4}}\mid =\mid z_2^{\frac{1}{4}}\mid ⩾\sqrt{17}>2$$$$\mathrm{otherwise}\mathrm{the}\mathrm{angles}\mathrm{and}\mathrm{the}\mathrm{absolutes}\mathrm{are}$$$$\mathrm{different}\mathrm{and}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{see}\mathrm{a}\mathrm{chance}\mathrm{the}\mathrm{sum}$$$$\mathrm{of}\mathrm{the}\mathrm{roots}\mathrm{might}\mathrm{be}\mathrm{real}$$

Answered by behi83417@gmail.com last updated on 04/Mar/20

$$17+\mathrm{x}=\mathrm{m},17-\mathrm{x}=\mathrm{n}$$$$\Rightarrow \{\begin{array}{l}\mathrm{m}+\mathrm{n}=34\\ {\mathrm{m}}^4+{\mathrm{n}}^4=2\end{array}\Rightarrow \{\begin{array}{l}\mathrm{let}:\mathrm{m}+\mathrm{n}=\mathrm{p},\mathrm{mn}=\mathrm{q}\\ \Rightarrow {\mathrm{m}}^4+{\mathrm{n}}^4={[{(\mathrm{m}+\mathrm{n})}^2-2\mathrm{mn}]}^2-{2\mathrm{m}}^2{\mathrm{n}}^2\\ \Rightarrow {\mathrm{m}}^4+{\mathrm{n}}^4={({\mathrm{p}}^2-2\mathrm{q})}^2-{2\mathrm{q}}^2={\mathrm{p}}^4-{4\mathrm{qp}}^2+{2\mathrm{q}}^2\end{array}$$$$\Rightarrow \{\begin{array}{l}\mathrm{p}=34\\ {\mathrm{p}}^4-{4\mathrm{qp}}^2+{2\mathrm{q}}^2=2\Rightarrow {34}^4-4\times {34}^2\times \mathrm{q}+{2\mathrm{q}}^2=2\end{array}$$$$\Rightarrow {\mathrm{q}}^2-2312\mathrm{q}+668167=0$$$$\Rightarrow \mathrm{q}=\frac{2312\pm \sqrt{{2312}^2-4\times 668167}}{2}\backsimeq 1973,338$$$$\Rightarrow \{\begin{array}{l}\mathrm{p}=34\\ \mathrm{q}=1973,338\end{array}$$$$\Rightarrow {\mathrm{z}}^2-34\mathrm{z}+(1973,338)=0\Rightarrow \mathrm{z}=\{\begin{array}{l}{\mathrm{z}}_{1,2}=34\pm 82\mathbf{i}\\ {\mathrm{z}}_{3,4}=34\pm 14\mathbf{i}\end{array}$$$$\Rightarrow \{\begin{array}{l}17+\mathrm{x}=34\pm 82\mathbf{i}\Rightarrow \mathrm{x}=17\pm 82\mathbf{i}\\ 17-\mathrm{x}=34\pm 82\mathbf{i}\Rightarrow \mathrm{x}=-17\mp 82\mathbf{i}\\ 17+\mathbf{x}=34\pm 14\mathbf{i}\Rightarrow \mathrm{x}=17\pm 14\mathbf{i}\\ 17-\mathbf{x}=34\pm 14\mathbf{i}\Rightarrow \mathrm{x}=-17\mp 14\mathbf{i}\end{array}$$

Commented by mr W last updated on 04/Mar/20

$$nicesolutionsir!$$

Commented by MJS last updated on 04/Mar/20

$$\mathrm{none}\mathrm{of}\mathrm{these}\mathrm{solve}\mathrm{the}\mathrm{given}\mathrm{equation}$$

Commented by MJS last updated on 04/Mar/20

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{this}\mathrm{has}\mathrm{any}\mathrm{solution}$$

Commented by behi83417@gmail.com last updated on 04/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{masters}:\mathrm{mrW}\mathrm{sir}\mathrm{and}$$$$\mathrm{MJS}\mathrm{sir}.$$$$\mathrm{any}\mathrm{way},\mathrm{i}\mathrm{think}\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{nice}\mathrm{method}$$$$\mathrm{for}\mathrm{solving}\mathrm{such}\mathrm{questions}.$$$$\mathrm{mybe}\mathrm{someone}\mathrm{likes}\mathrm{to}\mathrm{try}\mathrm{this}\mathrm{two}:$$$$1.\sqrt[4]{97-\mathbf{x}}+\sqrt[4]{\mathbf{x}}=5$$$$2.\sqrt{1-{\mathbf{x}}^2}={(\mathbf{a}-\sqrt{\mathbf{x}})}^2$$

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