First number = x
Second number = y
Case (i): Given that, the sum of both numbers is 27,
∴ x + y = 27
∴ x = 27 - y ------- (i)
Case (ii): Also, the product of both numbers is 182,
∴ xy = 182
∴ (27 - y)y = 182 { from (i) }
∴ 27y - y² = 182
∴ y² - 27y + 182 = 0
∴ y² - 13y - 14y + 182 = 0
∴ y(y - 13) - 14(y - 13) = 0
∴ (y - 13) (y - 14) = 0
Now,
y - 13 = 0 OR y - 14 = 0
∴ y = 13 ∴ y = 14
*Putting y = 13 in Eqⁿ(i)
∴ x = 27 - 13
∴ x = 14
*Putting y = 14 in Eqⁿ(i)
∴ x = 27 - 14
∴ x = 13
Hence,
The required two numbers are 13 and 14.
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Try This..
✒ Two numbers differ by 5 and their product is 104. Find the numbers.
✒The sum and product of two numbers are 16 and 55 respectively. Find the numbers.
✒ The sum of two numbers is 30 and their product is 221. Find the numbers.
✒ Find two numbers whose sum is 20 and product is 96.
❌ Common Mistakes
Forgetting the signs:▸ Always check the signs while expanding and factoring quadratic expressions.Incorrect factor pairs:▸ Make sure the numbers you pick for factoring actually multiply and add up to the required values.Skipping standard form:▸ Always write the quadratic in the ax² + bx + c = 0 format before solving.Not verifying:▸ Once you get the values, always plug them back to verify both the sum and product.
📝 Related Questions:-
- Ex 4.2 Q4 - Find two consecutive positive integers, sum of whose squares is 365.
- Ex 4.2 Q5 - The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
- Ex 4.2 Q6 - A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Queries Solved:-
Class 10 Ex 4.2
Ex 4.2 Q3 Class 10
Class 10 Ex 4.2 Q3
Class 10 Chap 4 Ex 4.2 Q3
Class 10 Quadratic Equations
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