Let the tens place digit be x and the
units place digits be y
so by the condition given in the
question
x+y=13. ----1
x^2+y^2=89
by squaring both the sides of equation
1 we get
x^2+2xy+y^2=169
89+2xy=169
2xy=169-89
xy=40
so
(x-y)^2=x^2-2xy +y^2
=89-80
=9
(x-y)^2=9
by taking square root of both the sides
we get
x-y=3. ---3
by adding equation 1 and 3 we get
2x=16
x=8
y=5
so the number is 85
Let the small no. be x and the big no be y
x+y=13...(1)
x^2+y^2=89...(2)
x=13-y
sub x in equ (2)
(13-y) (13-y) + y^2=89
169-13y-13y+y^2+y^2=89
2y^2-26y=-80
y^2-13y=-40
y^2-13y+40=0
Let the two digit be x and y respectively
x+y=13
x^2+y^2=89
frm eqn 1 y= 13_x
subtituting
x^2+(13_x)(13_×)=89
x^2+169_13x_13x+x^2=89
x^2+x^2_13x_13x+169=89
2x^2_26x+80=0
simplifying
x^2_13x+40=0
using _b+or _(b^2_4ac)/2a
=_ _ +13 square of + or _ (169_4×1×40)/2×1
=13 + or _ (9)/2
13+3/2=0 or 13_3/2=0
x=16/2=0 or 10/2 =0
x=4 or x=5
for x=4 ,4+y=13 y=13_4 y=11
for x=5, 5+y=13,y=13_5=8 y=8
1 month ago
units place digits be y
so by the condition given in the
question
x+y=13. ----1
x^2+y^2=89
by squaring both the sides of equation
1 we get
x^2+2xy+y^2=169
89+2xy=169
2xy=169-89
xy=40
so
(x-y)^2=x^2-2xy +y^2
=89-80
=9
(x-y)^2=9
by taking square root of both the sides
we get
x-y=3. ---3
by adding equation 1 and 3 we get
2x=16
x=8
y=5
so the number is 85