Quadratic Equations | Exercise 4.2

Question 4

Find two consecutive positive integers, sum of whose squares is 365.

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Solution
Understand the Question
  • Let two consecutive positive integers be xx and x+1x + 1.
  • The sum of their squares is given as 365365, which gives the equation x2+(x+1)2=365x^2 + (x + 1)^2 = 365.
  • We simplify this into a standard quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, solve for xx by factorisation, and select the positive integer root.

Step 1 · Formulate the Quadratic Equation

Let the two consecutive positive integers be xx and x+1x + 1.

According to the question

x2+(x+1)2=365x^2 + (x + 1)^2 = 365

Expand and simplify

x2+(x2+2x+1)=3652x2+2x+1=3652x2+2x+1−365=02x2+2x−364=0\begin{aligned} x^2 + (x^2 + 2x + 1) &= 365 \\ 2x^2 + 2x + 1 &= 365 \\ 2x^2 + 2x + 1 - 365 &= 0 \\ 2x^2 + 2x - 364 &= 0 \end{aligned}

Dividing the entire equation by 22

x2+x−182=0x^2 + x - 182 = 0

Step 2 · Solve the Quadratic Equation

Factorise by splitting the middle term (since 14×(−13)=−18214 \times (-13) = -182 and 14−13=114 - 13 = 1)

x2+14x−13x−182=0x(x+14)−13(x+14)=0(x−13)(x+14)=0\begin{aligned} x^2 + 14x - 13x - 182 &= 0 \\ x(x + 14) - 13(x + 14) &= 0 \\ (x - 13)(x + 14) &= 0 \end{aligned} x−13=0orx+14=0x - 13 = 0 \quad \text{or} \quad x + 14 = 0 x=13orx=−14x = 13 \quad \text{or} \quad x = -14

Since xx must be a positive integer, reject x=−14x = -14. Therefore, x=13x = 13.

The next consecutive integer is

13+1=1413 + 1 = 14

Verification: 132+142=169+196=36513^2 + 14^2 = 169 + 196 = 365.

Answer

13 and 14

Common Mistakes
  • Accepting the Negative Root: Forgetting that the question specifies positive integers and mistakenly including −14-14.
  • Expansion Error: Incorrectly expanding (x+1)2(x + 1)^2 as x2+1x^2 + 1 instead of x2+2x+1x^2 + 2x + 1.
  • Consecutive Integer Setup: Assuming consecutive integers are xx and x+2x + 2 (which applies to consecutive even or odd integers, not consecutive integers).

More questions in Exercise 4.2

Q1

Find the roots of the following quadratic equations by factorisation:

(i) x2−3x−10=0x^2 - 3x - 10 = 0

(ii) 2x2+x−6=02x^2 + x - 6 = 0

(iii) 2x2+7x+52=0\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0

(iv) 2x2−x+18=02x^2 - x + \dfrac{1}{8} = 0

(v) 100x2−20x+1=0100x^2 - 20x + 1 = 0

Q2

Solve the problems given in Example 1.

Q3

Find two numbers whose sum is 27 and product is 182.

Q4

Find two consecutive positive integers, sum of whose squares is 365.

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.

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