Paul is x years old and his father's age is twice the square of Paul's age. Ten years hence, father's age will be four times Pauls's age. Find their present ages.
Answers
Answer:
Paul present age = 5 years
Paul's father present age = 50 yrs
Step-by-step explanation:
Paul age = x
Paul's father age = 2x^2
After 10 yrs
Paul's age = x+10
Pauld father age = 2x^2 + 10
According to problem
after ten yrs,
Paul's father age = 4 × Paul's age
therefore
2x^2 + 10 = 4(x+10)
2x^2+ 10 = 4x+40
2x^2 - 4x - 30 = 0
x^2 -2x - 15 = 0
After factorising
x^2+3x-5x-15=0
x(x+3) -5(x+3)= 0
(x-5)(x+3)= 0
x=5
Paul's present age = 5 yrs
Paul's father present age = 50 yrs
Answer:
paul age 5 yr
Paul's father age 50 yr
Step-by-step explanation:
Paul age =x
Paul's father age= 2(x)^2 =>2x^2
after 10 years
paul age= x+10
Paul's father age = 2x^2+10
according to question after 10 years father age is four times of Paul
=>4(x+10)= 2x^2+10
=> 4x+40= 2x^2+10
=> 2x^2-4x+10-40=0
=> 2x^2-4x-30=0
divide whole equation by 2 we get
x^2-2x-15=0
=> x^2-5x+3x-15=0
=> x(x+3)-5(x+3)=0
=>(x+3)(x-5)=0
=> x=-3 and x=5
hence age never be negative so we take the value of x is positive 5 yr
=> paul present age is 5 and Paul's father age is 2(5*5)= 50 yr