Two years ago, my age was three times the square of my daughter 's age.In three years from now my age would be four times my daughter 's age.Find our present age.
Answers
Answered by
40
Let the daughters age be (2 yrs ago) = x
let the mothers age be (2 yrs ago) = 3x^{2}
present age of daughter = x+2
present age of mothers = 3x^{2}+2
After 3 yrs;
( 3x^{2}+2 +3) = 4(x+2+3)
3x^{2}+5 = 4x+20
3x^{2} - 4x-15 = 0
(x-3)(3x+5) = 0
x =3 AND other is rejected
y-2 = x
y-2 = 3
y = 5
let the mothers age be (2 yrs ago) = 3x^{2}
present age of daughter = x+2
present age of mothers = 3x^{2}+2
After 3 yrs;
( 3x^{2}+2 +3) = 4(x+2+3)
3x^{2}+5 = 4x+20
3x^{2} - 4x-15 = 0
(x-3)(3x+5) = 0
x =3 AND other is rejected
y-2 = x
y-2 = 3
y = 5
Answered by
0
Step-by-step explanation:
Two years ago. (x + 2) = (3x²+2) -
3 years from now
(30xx ((+2+3) = (3x² +2+3)
y(x + 5) = (3x ^ 2 + 5)
4x + 20 = 3x ^ 2 + 5
3x^ 2 +5=4.x+<0
3x ^ 2 + 5 - 4x + 20 3x ^ 2 - 4x - 15
3x ^ 2 - 9x + 5x - 15
3x(x - 3) + 5(x - 3) = 0
(3)(5)(x - 3) = 0 x = .5/3 x = 3 - 2
:ار
(B(3) ²+2) - ( JL² +2) (369+2) -) <3x²+2
2772
29
Branist
years and son
(x + 2) (3+2)
umbe
lenon
ence
nall
m
tern
er
2x
qu
n
Mark me
PLZ2
Attachments:
Similar questions