x-3-4-7-4-x-

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

$$\left({x}−\mathrm{3}\right)^{\mathrm{4}} =\mathrm{7}^{\mathrm{4}} \\ $$$${x}=? \\ $$

Answered by Ismoiljon_008 last updated on 19/Dec/24

∣ x − 3 ∣ = 7 x − 3 = 7 x = 10 ✓ x − 3 = − 7 x = − 4 ✓ Answer: x_1 = 10 ; x_2 = − 4

$$\:\:\:\mid\:{x}\:−\:\mathrm{3}\:\mid\:=\:\mathrm{7} \\ $$$$\:\:\:{x}\:−\:\mathrm{3}\:=\:\mathrm{7} \\ $$$$\:\:\:{x}\:=\:\mathrm{10}\:\:\checkmark \\ $$$$\:\:\:{x}\:−\:\mathrm{3}\:=\:−\:\mathrm{7} \\ $$$$\:\:\:{x}\:=\:−\:\mathrm{4}\:\:\checkmark \\ $$$$\:\:\:\mathbb{A}{nswer}:\:\:{x}_{\mathrm{1}} \:=\:\mathrm{10}\:;\:\:{x}_{\mathrm{2}} \:=\:−\:\mathrm{4} \\ $$

Answered by ppch145 last updated on 19/Dec/24

To solve for x, we can start by taking the fourth root of both sides of the equation: (x − 3)^(4 ×(1/4) ) = 7^(4 ×(1/4)) The powers cancel out, leaving us with x − 3 = ±7 Since the fourth root can have both positive and negative values, we have two possible equations: x − 3 = 7 or x − 3 = −7 Solving for x in both equations: x = 7+ 3 or x = −7 + 3 x = 10 or x = −4 So, the values of x are: 10 and −4.

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

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 { ((x−10=0⇒x=10)),((x+4=0⇒x=−4)),(((x−3)^2 +7^2 =0⇒(x−3)^2 =−49^★ )) :} ^★ x−3=±7i⇒x=3±7i

$$\left({x}−\mathrm{3}\right)^{\mathrm{4}} =\mathrm{7}^{\mathrm{4}} \:;\:\:{x}=? \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{4}} −\mathrm{7}^{\mathrm{4}} =\mathrm{0} \\ $$$$\left\{\left({x}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} \right\}\left\{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \right\}=\mathrm{0} \\ $$$$\left\{\left({x}−\mathrm{3}\right)−\mathrm{7}\right\}\left\{\left({x}−\mathrm{3}\right)+\mathrm{7}\right\}\left\{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \right\}=\mathrm{0} \\ $$$$\begin{cases}{{x}−\mathrm{10}=\mathrm{0}\Rightarrow{x}=\mathrm{10}}\\{{x}+\mathrm{4}=\mathrm{0}\Rightarrow{x}=−\mathrm{4}}\\{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} =\mathrm{0}\Rightarrow\left({x}−\mathrm{3}\right)^{\mathrm{2}} =−\mathrm{49}^{\bigstar} }\end{cases} \\ $$$$\:^{\bigstar} {x}−\mathrm{3}=\pm\mathrm{7}{i}\Rightarrow{x}=\mathrm{3}\pm\mathrm{7}{i} \\ $$

Answered by MrGaster last updated on 19/Dec/24

(x−3)^4 =(7)^4 (((x−3)^4 ))^(1/4) =(7^4 )^(1/4) x−3=±7 x=3±7 x−3+7,x−3−7 x=10 or x=−4

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

Facebook Tweet Pin

Leave a Reply Cancel reply