Question

how can be solved this for Y ?

Answer 1

\[\sqrt{7}(p - \sqrt{p}) /2(y^2 -p) = (y -\sqrt{p})^2 \]

Answer 2

my work

restriction y^2 -p not equal 0 so y^2 not equal p so y not equal +/- sqrt p

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jhonyy9
\[\sqrt{7}(p - \sqrt{p}) /2(y^2 -p) = (y -\sqrt{p})^2 \]
\(\color{#0cbb34}{\text{End of Quote}}\)
\[\sqrt{7}(p-\sqrt{p})=2(y-\sqrt{p})^{2}(y ^{2}- p)\]
\[2(y ^{2 }-2\sqrt{p}y +p)(y ^{2}-p) - \sqrt{7}(p - \sqrt{p}) = 0 \]

Answer 4

\[2(y ^{4} - 2\sqrt{p}y ^{3} +py ^{2 }-py ^{2} +2p \sqrt{p}y -p ^{2}) - \sqrt{7}(p - \sqrt{p}) = 0\]

\[y ^{4} - 2\sqrt{p}y ^{3}+ 2p \sqrt{p}y- p ^{2} = \sqrt{7}(p - \sqrt{p})/2\]

Answer 5

so from this can i getting anything clearly way to solve it ? or i ve tought it rewriting this equation in x

given that \[y^2 = p +x^2 \]

Answer 6

so i ve got

\[\sqrt{7}(p - \sqrt{p}) /2(p +x ^2 -p) = ( \sqrt{p +x ^{2 }}- \sqrt{p})^{2}\]

\[\sqrt{7}(p - \sqrt{p}) / 2x ^{2} = p +x ^{2} - 2\sqrt{p(p+x ^{2})} - p\]

\[\sqrt{7}(p - \sqrt{p}) = 2x ^{2}(x ^{2}- 2\sqrt{p(p+x ^{2})}\]

\[2x ^{2}(x ^{2}-2\sqrt{p(p+x ^{2})}) = \sqrt{7}(p-\sqrt{p})\]

\[2x ^{4} -4x ^{2}\sqrt{p(p+x ^{2})} - \sqrt{7}(p - \sqrt{p}) = 0\]

i ve continued from this with note x^2 = t but where i ve arrived was very confusably with t^4 and t^3 and t^2 and t

so any idea will be more apreciated - thank you in advance

Answer 7

@Vocaloid
@Angle