Question

Find two consecutive positive odd numbers such that the sum of whose squares is 802.

Answer 1

Hint: In such a type of question, we have to assume the two numbers let a and b. Here the required numbers are consecutive positive odd numbers and we know that difference between any two odd numbers is 2. So we can simply write $b = a + 2$. Given that the sum of squares of two numbers is 802. So, we can write as ${a^2} + {b^2} = 802$. Now putting $b = a + 2$, we can simplify the equation in one variable as ${a^2} + {(a + 2)^2} = 802$. Simplifying and solving the equation by use of quadratic formula $a = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$ we will get value of a and using $b = a + 2$, we will get the value of b. These two are our required consecutive positive odd numbers.

Complete step-by-step answer:
We have to find two consecutive positive odd numbers such that the sum of whose squares is 802.
Let’s assume the two required consecutive positive odd numbers as a and b.
Generally we know that the difference between any two odd numbers is 2. So we can simply write the second number (b) as first number (a) + 2.
So, $b = a + 2$
Here the sum of squares of both numbers is 802.
So, ${a^2} + {b^2} = 802$
Now putting $b = a + 2$ in above equation.
So, ${a^2} + {(a + 2)^2} = 802$
Using the formula of ${(x + y)^2}$, ${(x + y)^2} = {x^2} + {y^2} + 2xy$,
Expanding ${(a + 2)^2} = {a^2} + {2^2} + 2(a)(2)$
So, ${(a + 2)^2} = {a^2} + 4 + 4a$
Putting the value of ${(a + 2)^2}$ in above equation,
$\Rightarrow$${a^2} + {a^2} + 4 + 4a = 802$
$\Rightarrow$$2{a^2} + 4a + 4 = 802$
Dividing on both side of equation by 2,
$\Rightarrow$$\dfrac{{2{a^2} + 4a + 4}}{2} = \dfrac{{802}}{2}$
$\Rightarrow$$\dfrac{{2({a^2} + 2a + 2)}}{2} = 401$
$\Rightarrow$\[{a^2} + 2a + 2 = 401\]
Taking all terms on left side of equation \[{a^2} + 2a + 2 - 401 = 0\]
So, \[{a^2} + 2a - 399 = 0\]
To solve the above quadratic equation, we have to use a quadratic formula.
For a quadratic equation in form of $A{a^2} + Ba + C = 0$, where A, B and C are called coefficients of the quadratic equation.
The roots of quadratic equation ${a_1}$and ${a_2}$ can be obtained by using quadratic formula which is $a = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$

Comparing our quadratic equation with the general form of the quadratic equation we will get $A = 1$, $B = 2$ and $C = - 399$.
Now roots can obtained from $a = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$
Putting value of A,B and C,
$\Rightarrow$$a = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4(1)( - 399)} }}{{2(1)}}$
$\Rightarrow$$a = \dfrac{{ - 2 \pm \sqrt {4 + 1596} }}{2}$
Simplifying, $a = \dfrac{{ - 2 \pm \sqrt {1600} }}{2}$
$\Rightarrow$$a = \dfrac{{ - 2 \pm 40}}{2}$
So root ${a_1} = \dfrac{{ - 2 + 40}}{2}$ and ${a_2} = \dfrac{{ - 2 - 40}}{2}$
Simplifying ${a_1} = \dfrac{{ - 2 + 40}}{2} = \dfrac{{38}}{2} = 19$ and ${a_1} = \dfrac{{ - 2 - 40}}{2} = \dfrac{{ - 42}}{2} = - 21$
So our required number will be either 19 or -21. But the given condition is the number shall be positive so -21 is not possible.
Hence $a = 19$
Now $b = a + 2 = 19 + 2 = 21$

So, the required two consecutive positive odd numbers are 19 and 21.

Note: This type of question can be solved by assuming the numbers and converting all assumed numbers in single variable form. Please look at the table below for number assumptions for given conditions.

Sr. no.	Given condition for number	Assumption
1	If two consecutive odd/even numbers	a, a+2
2	If two consecutive numbers	a, a+1
3	If three consecutive numbers	a-1, a, a+1
4	If three consecutive odd/even numbers	a-2, a, a+2
5	If four consecutive odd/even numbers	a-3, a-1, a+1, a+3
6	If four consecutive numbers	a-1, a, a+1, a+2