Menu Close

Question-197452




Question Number 197452 by sonukgindia last updated on 18/Sep/23
Answered by Frix last updated on 18/Sep/23
t=5x^2 −2x−7  (√(t+4))−2=((2t))^(1/3)   Obviously t=−4∨t=0  t=−4 ⇒ x=−(3/5)∨x=1  t=0 ⇒ x=−1∨x=(7/5)  Less but still obvious t=32  ⇒ x=−((13)/5)∨x=3  We have  x∈{−((13)/5), −1, −(3/5), 1, (7/5), 3}
$${t}=\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{7} \\ $$$$\sqrt{{t}+\mathrm{4}}−\mathrm{2}=\sqrt[{\mathrm{3}}]{\mathrm{2}{t}} \\ $$$$\mathrm{Obviously}\:{t}=−\mathrm{4}\vee{t}=\mathrm{0} \\ $$$${t}=−\mathrm{4}\:\Rightarrow\:{x}=−\frac{\mathrm{3}}{\mathrm{5}}\vee{x}=\mathrm{1} \\ $$$${t}=\mathrm{0}\:\Rightarrow\:{x}=−\mathrm{1}\vee{x}=\frac{\mathrm{7}}{\mathrm{5}} \\ $$$$\mathrm{Less}\:\mathrm{but}\:\mathrm{still}\:\mathrm{obvious}\:{t}=\mathrm{32} \\ $$$$\Rightarrow\:{x}=−\frac{\mathrm{13}}{\mathrm{5}}\vee{x}=\mathrm{3} \\ $$$$\mathrm{We}\:\mathrm{have} \\ $$$${x}\in\left\{−\frac{\mathrm{13}}{\mathrm{5}},\:−\mathrm{1},\:−\frac{\mathrm{3}}{\mathrm{5}},\:\mathrm{1},\:\frac{\mathrm{7}}{\mathrm{5}},\:\mathrm{3}\right\} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *