Step-by-step explanation: First solution Let the ages of two people be a,ba,b . The sum of the ages of two friends is 20 years. → a+b=20a+b=20 …[Eqn. 1] Four years ago, the product of their ages was 48. → (a-4)(b-4)=48(a−4)(b−4)=48 …[Eqn. 2] Let’s find the product of their ages at now. Using both equations, → ab-4a-4b+16=48ab−4a−4b+16=48 → ab-4(a+b)+16=48ab−4(a+b)+16=48 → ab-80+16=48ab−80+16=48 → ab=112ab=112 …[Eqn. 3] Suppose a quadratic equation with their ages as zeros. By factor theorem, → (x-a)(x-b)=0(x−a)(x−b)=0 → x^2-(a+b)x+ab=0x 2 −(a+b)x+ab=0 We can use the information given in [Eqn. 2, 3]. → x^2-20x+112=0x 2 −20x+112=0 → D=20^2-4\times 112D=20 2 −4×112 → D=400-448 < 0D=400−448<0 Their ages are both imaginary numbers. So, this situation is not possible. Second solution Let the age of two people 4 years ago be a,ba,b . → a+b=12a+b=12 → ab=48ab=48 Using the same method as Sol. 1, we find two numbers a,ba,b . → x^2-12x+48=0x 2 −12x+48=0 → D=12^2-4\times48D=12 2 −4×48 → D=144-192 < 0D=144−192<0 This situation is not possible as their ages are both imaginary. Reply
Step-by-step explanation:
First solution
Let the ages of two people be a,ba,b .
The sum of the ages of two friends is 20 years.
→ a+b=20a+b=20 …[Eqn. 1]
Four years ago, the product of their ages was 48.
→ (a-4)(b-4)=48(a−4)(b−4)=48 …[Eqn. 2]
Let’s find the product of their ages at now. Using both equations,
→ ab-4a-4b+16=48ab−4a−4b+16=48
→ ab-4(a+b)+16=48ab−4(a+b)+16=48
→ ab-80+16=48ab−80+16=48
→ ab=112ab=112 …[Eqn. 3]
Suppose a quadratic equation with their ages as zeros. By factor theorem,
→ (x-a)(x-b)=0(x−a)(x−b)=0
→ x^2-(a+b)x+ab=0x
2
−(a+b)x+ab=0
We can use the information given in [Eqn. 2, 3].
→ x^2-20x+112=0x
2
−20x+112=0
→ D=20^2-4\times 112D=20
2
−4×112
→ D=400-448 < 0D=400−448<0
Their ages are both imaginary numbers. So, this situation is not possible.
Second solution
Let the age of two people 4 years ago be a,ba,b .
→ a+b=12a+b=12
→ ab=48ab=48
Using the same method as Sol. 1, we find two numbers a,ba,b .
→ x^2-12x+48=0x
2
−12x+48=0
→ D=12^2-4\times48D=12
2
−4×48
→ D=144-192 < 0D=144−192<0
This situation is not possible as their ages are both imaginary.
Answer:
(3x-2) +(2x+7) =0
3x-2-2x-7=0
x-9=0
x=9