Question 926309

if {{{(4P+1)^2=81}}} and {{{P>0}}},...solve for {{{P}}}

{{{(4P+1)^2-81=0}}}

{{{(4P+1)^2-9^2=0}}}

{{{((4P+1)-9)((4P+1)+9)=0}}}

{{{(4P+1-9)(4P+1+9)=0}}}

{{{(4P-8)(4P+10)=0}}}

solutions:

if {{{(4P-8)=0}}} =>{{{4P=8}}} =>{{{P=2}}}

if {{{(4P+10)=0}}} =>{{{4P=-10}}} =>{{{P=-5/2}}}


since given that {{{P>0}}}, we can use only {{{P=2}}} as a possible value for {{{P}}}