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Q4:
Find two consecutive positive integers, sum of whose squares is $365$.
Solution :
Given: Two consecutive positive integers. The sum of the squares of these integers is $365$.
To Find: The two consecutive positive integers.
Step 1: Defining the Variables
Let the first positive integer be $x$.
Since the integers are consecutive, the next integer must be $x + 1$.
[Assumption: $x$ is a positive integer, so $x > 0$].
Step 2: Formulating the Equation
According to the problem statement, the sum of the squares of these two integers is $365$.
Mathematically, this is expressed as:
$x^2 + (x + 1)^2 = 365$
Step 3: Expanding and Simplifying the Equation
Apply the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$ to expand $(x + 1)^2$:
$x^2 + (x^2 + 2x + 1) = 365$
Combine like terms:
$2x^2 + 2x + 1 = 365$
Subtract $365$ from both sides to set the quadratic equation to the standard form $ax^2 + bx + c = 0$:
$2x^2 + 2x + 1 - 365 = 0$
$2x^2 + 2x - 364 = 0$
Step 4: Simplifying the Quadratic Equation
Divide the entire equation by $2$ to simplify the coefficients:
$x^2 + x - 182 = 0$
Step 5: Solving by Factorization
We need to find two numbers that multiply to $-182$ and add to $1$ (the coefficient of $x$).
The factors of $182$ are: $1 \times 182, 2 \times 91, 7 \times 26, 13 \times 14$.
Since the product is negative and the sum is positive, we choose $14$ and $-13$.
$x^2 + 14x - 13x - 182 = 0$
Group the terms:
$x(x + 14) - 13(x + 14) = 0$
Factor out the common binomial $(x + 14)$:
$(x + 14)(x - 13) = 0$
Step 6: Finding the Roots
Set each factor to zero:
1) $x + 14 = 0 \implies x = -14$
2) $x - 13 = 0 \implies x = 13$
Step 7: Evaluating the Results
The problem specifies that the integers must be positive.
Therefore, we reject $x = -14$.
We accept $x = 13$.
Step 8: Determining the Consecutive Integer
If the first integer is $x = 13$, then the second consecutive integer is:
$x + 1 = 13 + 1 = 14$.
Verification:
$13^2 + 14^2 = 169 + 196 = 365$. The condition is satisfied.
Final Answer: The two consecutive positive integers are 13 and 14.
More Questions from Class 10 Mathematics Quadratic Equations EXERCISE 4.2
- Q1(i): Find the roots of the following quadratic equations by factorisation: (i) $x^2 – 3x – 10 = 0$
- Q1(ii): Find the roots of the following quadratic equations by factorisation: (ii) $2x^2 + x – 6 = 0$
- Q1(iii): Find the roots of the following quadratic equations by factorisation: (iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
- Q1(iv): Find the roots of the following quadratic equations by factorisation: (iv) $2x^2 – x + \frac{1}{8} = 0$
- Q1(v): Find the roots of the following quadratic equations by factorisation: (v) $100x^2 – 20x + 1 = 0$
- Q2: Solve the problems given below. Represent the following situations mathematically: (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs.750. We would like to find out the number of toys produced on that day.
- Q3: Find two numbers whose sum is $27$ and product is $182$.
- 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.
- 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.
CBSE Solutions for Class 10 Mathematics Quadratic Equations
Chapters in CBSE - Class 10 Mathematics
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