Question

4) Solve the following problems using quadratic equations
a) The difference of two positive integers is 3 and the sum of their squares is
117; find the numbers.
b) The product of two consecutive integers is 3906. Find the integers.

Answer 1

Answer :

(a) The required numbers are 6 and 9

(b) The required integers may be 62 and 63 or -62 or -63

Given :

For Question (a)

• The difference of two positive integer is 3
• The sum of their square is 117

For Question (b)

• The product of two consecutive integers is 3906

To Find :

• (a) The two positive integers or numbers
• (b) The integers

Solution :

Question : (a)

Let us consider the positive integers be x and x + 3

According to question :

$\sf x^{2} + (x + 3)^{2} = 117 \\\\ \sf \implies x^{2} + x^{2} + 6x + 9 - 117 = 0 \\\\ \sf \implies 2x^{2} + 6x -108 \\\\ \sf \implies 2x^{2} + 18x - 12x - 108 \\\\ \sf \implies 2x(x + 9) - 12(x + 9) = 0 \\\\ \sf \implies (2x - 12)(x + 9) = 0$

Therefore , we have :

$\sf 2x - 12=0 \: \: and \: \: x + 9 =0\\\\ \sf \implies x = 6 \: and \implies x = -9$

Since x is a positive integer so x≠-9 .

Thus the required positive integers are :

6 and 6+3

or 6 and 9

_________________________

Question (b)

Let us consider the consecutive integers be x and x + 1

According to Question :

$\sf \implies x(x+1) = 3906 \\\\ \sf \implies x^{2} + x - 3906 = 0 \\\\ \sf \implies x^{2} + 63x - 62x - 3906 = 0 \\\\ \sf \implies x(x + 63) - 62(x + 63) = 0 \\\\ \sf \implies (x +63)(x - 62) = 0$

Therefore ,

$\sf x + 63 = 0 \: \: and \: \: x - 62 = 0 \\\\ \sf \implies x = -63 \: \: and \implies x = 62$

Thus , required integers are :

-63 and -63 + 1 or 62 and 62+1

→ -63 and -62 or 62 and 63