SOLUTION: Let the daughter's age be x Let the father's age be 4x 3 years ago: Daughter's age = x-3 Father's age = 4x-3 Product of their ages:( x-3)(4x-3)=430 Opening the bracket and app

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Question 776905: Let the daughter's age be x
Let the father's age be 4x
3 years ago: Daughter's age = x-3
Father's age = 4x-3
Product of their ages:( x-3)(4x-3)=430
Opening the bracket and applying quadratic formula method, we shall have the roots as x= 12.3 and -8.6
So, taking the positive value of x= 12.3years.
Present ages:
Daughter's (x-3)years = 12.3-3 =9.30years (9 years 3 months)
Father's age: (4x-3)years= 4×12.3-3 = 46.21years (46 years 2 months 1day)
Checking:
9.30×46.21= 429.753 = 430 (approximated)
Found 4 solutions by psbhowmick, MathTherapy, Alan3354, sciencetechnologyproductex@yahoo.co.uk:
Answer by psbhowmick(878)   (Show Source): You can put this solution on YOUR website!
Let father's and daughter's age be x and y years respectively now.

Now:
xy = 430 ________ (1)

3 years before:
x - 3 = 4y
x = 4y + 3 ________ (2)

Substituting x from (2) into (1)



Solve this equation to find y
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

The discriminant -6871 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -6871 is + or - .

The solution is

Here's your graph:



Imaginary roots suggest that this is impossible. Check and ensure that the data for the question is correct.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
3 years ago a father was 4 times as old as his daughter is now. The product of their ages is 430. Calculate their present ages.

Let daughter's, and father's current ages be D, and F, respectively

Then: F - 3 = 4D ----- F = 4D + 3 ----- eq (i)

Also FD = 430 ------ eq (ii)

D(4D + 3) = 430 ----- Substituting for F in eq (ii)



Solve this to get daughter's age. Then find father's:

You should get:

Daughter's current age:

Father's current age:

You can do the check!!

Further help is available, online or in-person, for a fee, obviously. Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
3 years ago a father was 4 times as old as his daughter is now. The product of their ages is 430. Calculate their present ages.
---------------
F*D = 430
F-3 = 4(D-3)
F = 4D - 9
----
D*(4D-9) = 430
4D^2 - 9D - 430 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=6961 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 11.5540759418081, -9.30407594180808. Here's your graph:

----------------------
D = x
----
Not an integer solution.
Answer by sciencetechnologyproductex@yahoo.co.uk(1)   (Show Source): You can put this solution on YOUR website!
Let the daughter's age be x
Let the father's age be 4x
3 years ago: Daughter's age = x-3
Father's age = 4x-3
Product of their ages:( x-3)(4x-3)=430
Opening the bracket and applying quadratic formula method, we shall have the roots as x= 12.3 and -8.6
So, taking the positive value of x= 12.3years.
Present ages:
Daughter's (x-3)years = 12.3-3 =9.30years (9 years 3 months)
Father's age: (4x-3)years= 4×12.3-3 = 46.21years (46 years 2 months 1 week)
Checking:
9.30×46.21= 429.753 = 430 (approximated
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