
The sum of squares of two consecutive odd numbers is 650. Which following equations satisfy the given statement?
A. $ {x^2} + {x^2} + {2^2} = 650 $
B. $ {x^2} + 2x - 323 = 0 $
C. $ 2{x^2} + 4 = 650 $
D. None of these.
Answer
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Hint: Let one odd number of the two consecutive odd numbers be x. Every two consecutive odd numbers leave a difference of 2 when subtracted the smaller one from larger. So the consecutive odd number of x will be $ x + 2 $ . Square these two odd numbers x and $ x + 2 $ and find their sum. We will get an equation in terms of x and that will be our answer. The resulting equation must be quadratic.
Complete step-by-step answer:
We are given that the sum of squares of two consecutive odd numbers is 650.
We have to find the equation which satisfies the above given statement.
Let one odd number of the two consecutive odd numbers be x.
Then the consecutive odd number of x will be $ x + 2 $ as every two consecutive odd numbers have a difference 2 between them.
Square of x is $ {x^2} $ and square of $ x + 2 $ is $ {\left( {x + 2} \right)^2} $
Sum of the squares of these consecutive odd numbers is $ {x^2} + {\left( {x + 2} \right)^2} $ which is equal to 650 as given.
$ \Rightarrow {x^2} + {\left( {x + 2} \right)^2} = 650 $
$ \Rightarrow {x^2} + {x^2} + 2\left( x \right)\left( 2 \right) + {2^2} = 650 $ as $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $
$ \Rightarrow 2{x^2} + 4x + 4 = 650 $
$ \Rightarrow 2{x^2} + 4x = 650 - 4 = 646 $
$ \Rightarrow 2\left( {{x^2} + 2x} \right) = 646 $
$ \Rightarrow {x^2} + 2x = \dfrac{{646}}{2} = 323 $
Therefore, the equation is $ {x^2} + 2x = 323 $ which can also be written as $ {x^2} + 2x - 323 = 0 $
So, the correct answer is “Option B”.
Note: Difference between any two consecutive odd numbers is 2 whereas difference between any two consecutive even numbers is also 2. An odd number is a number which is not divisible by 2 whereas even number is a number which is divisible by 2. We can verify the equation we got by finding the values of x and the obtained values must leave a remainder 1 when divided by 2.
Complete step-by-step answer:
We are given that the sum of squares of two consecutive odd numbers is 650.
We have to find the equation which satisfies the above given statement.
Let one odd number of the two consecutive odd numbers be x.
Then the consecutive odd number of x will be $ x + 2 $ as every two consecutive odd numbers have a difference 2 between them.
Square of x is $ {x^2} $ and square of $ x + 2 $ is $ {\left( {x + 2} \right)^2} $
Sum of the squares of these consecutive odd numbers is $ {x^2} + {\left( {x + 2} \right)^2} $ which is equal to 650 as given.
$ \Rightarrow {x^2} + {\left( {x + 2} \right)^2} = 650 $
$ \Rightarrow {x^2} + {x^2} + 2\left( x \right)\left( 2 \right) + {2^2} = 650 $ as $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $
$ \Rightarrow 2{x^2} + 4x + 4 = 650 $
$ \Rightarrow 2{x^2} + 4x = 650 - 4 = 646 $
$ \Rightarrow 2\left( {{x^2} + 2x} \right) = 646 $
$ \Rightarrow {x^2} + 2x = \dfrac{{646}}{2} = 323 $
Therefore, the equation is $ {x^2} + 2x = 323 $ which can also be written as $ {x^2} + 2x - 323 = 0 $
So, the correct answer is “Option B”.
Note: Difference between any two consecutive odd numbers is 2 whereas difference between any two consecutive even numbers is also 2. An odd number is a number which is not divisible by 2 whereas even number is a number which is divisible by 2. We can verify the equation we got by finding the values of x and the obtained values must leave a remainder 1 when divided by 2.
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