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Quadratic Equations
EXERCISE 4.3
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Step-by-Step Solution

Key Concept: Form a system of equations using the given sum and product conditions, substitute one variable from the linear equation into the quadratic equation, and solve the resulting quadratic by finding its discriminant. If the discriminant is negative, the quadratic has no real roots, indicating the situation is impossible.
1. Let the present ages of the two friends be $x$ years and $y$ years.\
2. From the given sum: $$x + y = 20 \[4pt] \Rightarrow y = 20 - x.$$\
3. Four years ago their ages were $x-4$ and $y-4$. The product of those ages was 48: $$ (x-4)(y-4) = 48. $$\
4. Substitute $y = 20 - x$ into the product equation: $$ (x-4)\big[(20 - x)-4\big] = 48 \Rightarrow (x-4)(16 - x) = 48. $$\
5. Expand the left‑hand side: $$ (x-4)(16 - x) = 16x - x^{2} - 64 + 4x = -x^{2} + 20x - 64. $$\
6. Set the expanded expression equal to 48 and bring all terms to one side: $$ -x^{2} + 20x - 64 = 48 \Rightarrow -x^{2} + 20x - 112 = 0. $$\
7. Multiply by $-1$ to obtain the standard quadratic form: $$ x^{2} - 20x + 112 = 0. $$\
8. Compute the discriminant $D$: $$ D = b^{2} - 4ac = (-20)^{2} - 4\cdot1\cdot112 = 400 - 448 = -48. $$\
9. Since $D < 0$, the quadratic equation has no real solutions. Hence there are no real (and therefore no possible integer) ages $x$ and $y$ that satisfy both conditions.\
10. Conclusion: The described situation is not possible.

Correct Answer: No such ages exist; the situation is not possible.
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