Answer
Verified
460.5k+ views
Hint: Here, we need to find the value of \[8.5 \times 9.5\] using a suitable standard identity. We will rewrite the two numbers as a sum or difference of two numbers in such a way that the product can be found using one of the standard algebraic identities. Then, we will simplify the expression to find the required value.
Formula Used: The product of the sum of two number and the difference of two numbers can be calculated using the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
Complete step-by-step answer:
We can evaluate the given product using any of the two identities \[\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\] or \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
We will use the second identity to solve this problem.
First, we will rewrite the given numbers as the sum or difference of two numbers such that the identity is applicable.
We know that \[8.5\] is the difference of 9 and \[0.5\], and \[9.5\] is the sum of 9 and \[0.5\].
Therefore, substituting \[8.5 = 9 - 0.5\] and \[9.5 = 9 + 0.5\], we can rewrite the product as
\[ \Rightarrow 8.5 \times 9.5 = \left( {9 - 0.5} \right)\left( {9 + 0.5} \right)\]
Now, we know that the product of the sum of two number and the difference of two numbers can be calculated using the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
Substituting \[a = 9\] and \[b = 0.5\] in the algebraic identity, we get
\[ \Rightarrow \left( {9 - 0.5} \right)\left( {9 + 0.5} \right) = {9^2} - {\left( {0.5} \right)^2}\]
Simplifying the expression, we get
\[ \Rightarrow 8.5 \times 9.5 = 81 - 0.25\]
Subtracting \[0.25\] from 81, we get
\[ \Rightarrow 8.5 \times 9.5 = 80.75\]
Therefore, the value of the product \[8.5 \times 9.5\] is \[80.75\].
Note: We can also solve the problem using the identity \[\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\].
Substituting \[8.5 = 8 + 0.5\] and \[9.5 = 8 + 1.5\], we can rewrite the product as
\[ \Rightarrow 8.5 \times 9.5 = \left( {8 + 0.5} \right)\left( {8 + 1.5} \right)\]
Now, we will use the algebraic identity \[\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\].
Substituting \[x = 8\],\[a = 0.5\] and \[b = 1.5\] in the algebraic identity, we get
\[ \Rightarrow \left( {8 + 0.5} \right)\left( {8 + 1.5} \right) = {8^2} + \left( {0.5 + 1.5} \right)8 + \left( {0.5} \right)\left( {1.5} \right)\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 8.5 \times 9.5 = 64 + 2 \times 8 + 0.75\\ \Rightarrow 8.5 \times 9.5 = 64 + 16 + 0.75\end{array}\]
Adding the terms of the expression, we get
\[ \Rightarrow 8.5 \times 9.5 = 80.75\]
\[\therefore\] The value of the product \[8.5 \times 9.5\] is \[80.75\].
Formula Used: The product of the sum of two number and the difference of two numbers can be calculated using the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
Complete step-by-step answer:
We can evaluate the given product using any of the two identities \[\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\] or \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
We will use the second identity to solve this problem.
First, we will rewrite the given numbers as the sum or difference of two numbers such that the identity is applicable.
We know that \[8.5\] is the difference of 9 and \[0.5\], and \[9.5\] is the sum of 9 and \[0.5\].
Therefore, substituting \[8.5 = 9 - 0.5\] and \[9.5 = 9 + 0.5\], we can rewrite the product as
\[ \Rightarrow 8.5 \times 9.5 = \left( {9 - 0.5} \right)\left( {9 + 0.5} \right)\]
Now, we know that the product of the sum of two number and the difference of two numbers can be calculated using the algebraic identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\].
Substituting \[a = 9\] and \[b = 0.5\] in the algebraic identity, we get
\[ \Rightarrow \left( {9 - 0.5} \right)\left( {9 + 0.5} \right) = {9^2} - {\left( {0.5} \right)^2}\]
Simplifying the expression, we get
\[ \Rightarrow 8.5 \times 9.5 = 81 - 0.25\]
Subtracting \[0.25\] from 81, we get
\[ \Rightarrow 8.5 \times 9.5 = 80.75\]
Therefore, the value of the product \[8.5 \times 9.5\] is \[80.75\].
Note: We can also solve the problem using the identity \[\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\].
Substituting \[8.5 = 8 + 0.5\] and \[9.5 = 8 + 1.5\], we can rewrite the product as
\[ \Rightarrow 8.5 \times 9.5 = \left( {8 + 0.5} \right)\left( {8 + 1.5} \right)\]
Now, we will use the algebraic identity \[\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\].
Substituting \[x = 8\],\[a = 0.5\] and \[b = 1.5\] in the algebraic identity, we get
\[ \Rightarrow \left( {8 + 0.5} \right)\left( {8 + 1.5} \right) = {8^2} + \left( {0.5 + 1.5} \right)8 + \left( {0.5} \right)\left( {1.5} \right)\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 8.5 \times 9.5 = 64 + 2 \times 8 + 0.75\\ \Rightarrow 8.5 \times 9.5 = 64 + 16 + 0.75\end{array}\]
Adding the terms of the expression, we get
\[ \Rightarrow 8.5 \times 9.5 = 80.75\]
\[\therefore\] The value of the product \[8.5 \times 9.5\] is \[80.75\].
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE