[tex]\huge{ \underline{\boxed{\red{ \color{red}{\textsf {\textbf{An}{\color{blue}{\textbf {\textsf{sw}}{\color{green}{\textbf{\textsf{er}}}{\color{orange}{\textsf{\textbf{~:-}}}}}}}}}}}}}[/tex] let the required no. be x and y so according to the question => x+ y = 50 ——-(1) and => xy = 525 ——-(2) so put y = 525/x in eqn (1) we get , => x + 525/x = 50 => x^2 + 525 = 50x => x^2 -50x +525 = 0 => x^2 -35x -15x + 525 =0 => x(x-35) -15(x-35) =0 => (x-35)(x-15) = 0 => x-35 = 0 or x-15 = 0 => x = 35 , 15 # so the required number is 35 and 15 ______________________________ [tex]\sf\blue{hope \: this \: helps \: you!! \: }.[/tex] Reply
let two numbers be € and ₩ so €₩= 525 and €+₩= 50 so ₩=50-€ so €(50-€)=525 50€-€²-525=0 solving the quadratic.. €²-50€+525=0 €=50+-(2500-2100)½/2 €=50+-20/2 €=35 and 15 so the two natural numbers are 35 and 15. Reply
[tex]\huge{ \underline{\boxed{\red{ \color{red}{\textsf {\textbf{An}{\color{blue}{\textbf {\textsf{sw}}{\color{green}{\textbf{\textsf{er}}}{\color{orange}{\textsf{\textbf{~:-}}}}}}}}}}}}}[/tex]
let the required no. be x and y
so according to the question
=> x+ y = 50 ——-(1)
and => xy = 525 ——-(2)
so put y = 525/x in eqn (1)
we get ,
=> x + 525/x = 50
=> x^2 + 525 = 50x
=> x^2 -50x +525 = 0
=> x^2 -35x -15x + 525 =0
=> x(x-35) -15(x-35) =0
=> (x-35)(x-15) = 0
=> x-35 = 0 or x-15 = 0
=> x = 35 , 15
# so the required number is 35 and 15
______________________________
[tex]\sf\blue{hope \: this \: helps \: you!! \: }.[/tex]
let two numbers be € and ₩
so
€₩= 525 and
€+₩= 50
so ₩=50-€
so €(50-€)=525
50€-€²-525=0
solving the quadratic..
€²-50€+525=0
€=50+-(2500-2100)½/2
€=50+-20/2
€=35 and 15
so the two natural numbers are 35 and 15.