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Question Number 214785 by mathlove last updated on 19/Dec/24

$${(x-3)}^4=7^4$$$$x=?$$

Answered by Ismoiljon_008 last updated on 19/Dec/24

$$\mid x-3\mid =7$$$$x-3=7$$$$x=10✓$$$$x-3=-7$$$$x=-4✓$$$$\mathbb{A}nswer:x_1=10;x_2=-4$$

Answered by ppch145 last updated on 19/Dec/24

$$\mathrm{To}\mathrm{solve}\mathrm{for}\mathrm{x},\mathrm{we}\mathrm{can}\mathrm{start}\mathrm{by}$$$$\mathrm{taking}\mathrm{the}\mathrm{fourth}\mathrm{root}\mathrm{of}\mathrm{both}$$$$\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{equation}:$$$$$$$${(\mathrm{x}-3)}^{4\times \frac{1}{4}}=7^{4\times \frac{1}{4}}$$$$$$$$\mathrm{The}\mathrm{powers}\mathrm{cancel}\mathrm{out},\mathrm{leaving}$$$$\mathrm{us}\mathrm{with}\mathrm{x}-3=\pm 7$$$$$$$$\mathrm{Since}\mathrm{the}\mathrm{fourth}\mathrm{root}\mathrm{can}\mathrm{have}$$$$\mathrm{both}\mathrm{positive}\mathrm{and}\mathrm{negative}$$$$\mathrm{values},\mathrm{we}\mathrm{have}\mathrm{two}\mathrm{possible}$$$$\mathrm{equations}:$$$$$$$$\mathrm{x}-3=7\mathrm{or}\mathrm{x}-3=-7$$$$$$$$\mathrm{Solving}\mathrm{for}\mathrm{x}\mathrm{in}\mathrm{both}\mathrm{equations}:$$$$$$$$\mathrm{x}=7+3\mathrm{or}\mathrm{x}=-7+3$$$$$$$$\mathrm{x}=10\mathrm{or}\mathrm{x}=-4$$$$$$$$\mathrm{So},\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{are}:10\mathrm{and}-4.$$$$$$$$$$$$$$$$$$

Answered by Rasheed.Sindhi last updated on 19/Dec/24

$${(x-3)}^4=7^4;x=?$$$${(x-3)}^4-7^4=0$$$$\{{(x-3)}^2-7^2\}\{{(x-3)}^2+7^2\}=0$$$$\{(x-3)-7\}\{(x-3)+7\}\{{(x-3)}^2+7^2\}=0$$$$\{\begin{array}{l}x-10=0\Rightarrow x=10\\ x+4=0\Rightarrow x=-4\\ {(x-3)}^2+7^2=0\Rightarrow {(x-3)}^2=-{49}^{★}\end{array}$$$${}^{★}x-3=\pm 7i\Rightarrow x=3\pm 7i$$

Answered by MrGaster last updated on 19/Dec/24

$${(x-3)}^4={\left(7\right)}^4$$$$\sqrt[4]{{(x-3)}^4}=\sqrt[4]{7^4}$$$$x-3=\pm 7$$$$x=3\pm 7$$$$x-3+7,x-3-7$$$$x=10\mathrm{or}x=-4$$

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