Hi,

a woman is 4 times older than her daughter's age , 

6 years ago the product of their ages was 136.

Question states***  6 years ago...

 (x-6)(4x-6) = 136

  4x^2 -30x + 36 = 136

  4x^2 -30x - 100 = 0

  2x^2 - 15x - 50 = 0







    x = 10,  throwing out the negative solution for age

 daughter is 10 and her mother is 40   



and... 6 years ago...checking our answer

 



 

Answer by Maths68(1474)   (Show Source): You can put this solution on YOUR website!
 Let

Present age of woman = w years old

Present age of daughter = d years old





woman is 4 times older than her daughter

w=4d............(1)

6 years ago 

woman was = w-6 years old

daughter was = d-6

then

product of their ages was 136

(w-6)(d-6)=136

substitute the value of from (1) to above equation

(4d-6)(d-6)=136

4d^2-24d-6d+36=136

4d^2-30d+36=136

2(2d^2-15d+18)=136

Divide by 2 both sides

2(2d^2-15d+18)/2=136/2

2d^2-15d+18=68

2d^2-15d+18-68=0

2d^2-15d-50=0



 
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 : .

  

  Discriminant d=625 is greater than zero. That means that there are two solutions: .

  

    

    

    

    Quadratic expression  can be factored:

  

  Again, the answer is: 10, -2.5.
Here's your graph:





d=10 or d=-2.5 (unacceptable)

so

d=10

Put the value of d in (1)

w=4d

w=4(10)

w=40





Present age of woman = w = 40 years old

Present age of daughter = d = 10 years old

 

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