# The sum of ages of Lisa and her mother is 45. The product of their ages is 126. Find their ages.

There are two ways to go about this problem. The first is to take all the denominators of 126 and experiment with adding them together until you get two that add to 45. In this case, start with dividing by two and work your way up:

2*63 = 123, but 2 + 63 = 65

3*42 = 126, and 3 +42 = 45  - bingo!

The other method is to set up a set of equations:

x + y =45

x * y = 126

x=45-y

(45-y)*y=126

-y^2 + 45y = 126

0 = y^2-45y +126

Then use the quadratic formula to get the solution. The quadratic formula, as you may recall, is

x = [ -b +- rad(b^2-4ac) ] / 2a

where the expression above is in the format:

ax^2 + bx +c = 0

In this case, a=1, b=-45, and c=126.

I'll leave the chugging for exercise.

epollock | Student

Solution:

Let Lisa’s age = x years

Her mother’s age = (45 – x) years

The product of their ages = x(45 –x) = 45x –x2

It is given that the product of their ages is 126.45x – x2 = 126

or x2 – 45x + 126 = 0

or x2 – 42x – 3x + 126 = 0

or x(x – 42) – 3(x – 42) = 0

or (x – 42) (x – 3) = 0

Either x – 42 = 0 or x – 3 = 0

x = 42, or x = 3

We reject x = 42 because Lisa’s age cannot be greater than her mother’s age.

x = 3

Lisa’s age is 3 years and her mother’s age = 45 – 3 = 42 years.

clash1016 | Student

L(isa) + M(om) = 45 and LM=126.

guess and check.... let L=5.  so M would = 40, but 40 X 5 = 200, so we know 5 is too hight.  Try L=3.  If L=3, then M=42.  42 X 3= 126, so Lisa is 3 and Mom is 42.

giorgiana1976 | Student

This could be a practical example of Viete's relationships between coefficients and roots of an second grade equation.

We could find a second grade equation using this relation:

X^2 - S*X + P=0, where S=x1+x2 (the sum between the 2 roots of eq.) and P=x1*x2 (product of the 2 roots of the same eq.)

So,the sum of ages of Lisa and her mother is 45 and let's suppose that the mother's age is x1 and Lisa's age is x2.

S=x1+x2=45

The product of their ages is 126.

P=x1*x2=126

Now, all we have to do is to create the second grade equation:

X^2-45*X+126=0

x1={-(-45)+sqrt[(-45)^2-4*1*126]}/2*1

x1= [45+sqrt(2025-504)]/2

x1= [45+sqrt(1521)]/2

x1= (45+39)/2

x1= 42

x2={-(-45)-sqrt[(-45)^2-4*1*126]}/2*1

x2=(45-39)/2

x2=3

So, Lisa's mother age is 42 and Lisa's age is 3.

neela | Student

Let Lisa and her mother's ages be respectively L and M years.

Therefore by data the sum of their ages, M+L = 45 and

the product of their ages is 126 or ML = 126.

We know that (A-B)^2 = (A+B)^2-4AB is an identy. So,

A-B = sqrt{(A-B)^2 - \$AM}. Let us yse this identity to find M - L.

Therefore, (M - L) = sqrt {(M+L)^2 - 4ML} = sqrt{45^2-4*126) = sqrt 1521 = 39.

So,

M + L = 45.......... .(1) and

M - L = 39............(2)

Eq(1) + Eq(2) gives: 2M= 45+39 = 84. Or M = 84/2 = 42 years is the age of the Mother of Lisa.

Eq(1) - Eq(2) gives: = 2M = 45 - 39 =  6 Or M = 45 -39 = 6 Or L= 6/2 =3 years is Lisa's age.

Tally:

M+l = 42+3 =45 and ML = 42*3 =126.

krishna-agrawala | Student

Let us say that Lisa's age ix x and her mothers age is y. Then as per the data given in the question:

x + y = 45   ...   (1)

x*y = 126   ...   (2)

Dividing both sides of equation 2 by y we get:

x = 126/y

Substituting this value of x in equation 1 we get:

126/y + y = 45

multiplying both side of above equation by y and taking all the terms on one side we get:

y^2 - 45y + 126 = 0

Therefore: y^2 - 3y - 42y + 126 = 0

Therefore: y(y - 3) - 42(y - 3) = 0

Therefore: (y - 3)*(y - 42) = 0

Therefore y = 3 or y = 42

Out of these two values 3 represents age of Lisa and 42 represents age of her mother.