
If \[{{\left( 64 \right)}^{2}}-{{\left( 36 \right)}^{2}}=20x\] then the value of \[x\] is:
(a) 70
(b) 120
(c) 180
(d) 140
(e) None of the aboHv
Answer
506.7k+ views
Hint: We solve this problem by using the simple formula of algebra.
We have the formula for the difference of the square of two numbers that is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
By using this formula for the given equation and by cross multiplying the terms of the given equation we can calculate the required value.
Complete step by step answer:
We are given that the equation as
\[{{\left( 64 \right)}^{2}}-{{\left( 36 \right)}^{2}}=20x\]
Now, let us use the formula of difference of squares of two numbers
We know that the formula for difference of square of two numbers that is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
Now, by using this formula to given equation we get
\[\begin{align}
& \Rightarrow \left( 64+36 \right)\left( 64-36 \right)=20x \\
& \Rightarrow 100\times 28=20x \\
\end{align}\]
We know that changing the multiplication from one side of equation to other side it will become division.
Now, by interchanging all the terms except the variable that is taking all terms from RHS to LHS except the variable we get
\[\begin{align}
& \Rightarrow \dfrac{100\times 28}{20}=x \\
& \Rightarrow x=5\times 28 \\
& \Rightarrow x=140 \\
\end{align}\]
Therefore we can conclude that the value of \[x\] is 140.
So, option (d) is the correct answer.
Note:
We can solve this problem without using the formula of algebra.
We are given with the equation that is
\[{{\left( 64 \right)}^{2}}-{{\left( 36 \right)}^{2}}=20x\]
We know that square of a number is multiplying the number with itself that is
\[{{n}^{2}}=n\times n\]
Now, by using the above formula to given equation we get
\[\Rightarrow \left( 64\times 64 \right)-\left( 36\times 36 \right)=20x\]
Now, by applying the required operations in the above equation we get
\[\begin{align}
& \Rightarrow 4096-1296=20x \\
& \Rightarrow 2800=20x \\
\end{align}\]
Now, by cross multiplying the terms that is taking all terms from RHS to LHS except the variable we get
\[\begin{align}
& \Rightarrow \dfrac{2800}{20}=x \\
& \Rightarrow x=\dfrac{280}{2} \\
& \Rightarrow x=140 \\
\end{align}\]
Therefore we can conclude that the value of \[x\] is 140.
So, option (d) is correct answer
We have the formula for the difference of the square of two numbers that is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
By using this formula for the given equation and by cross multiplying the terms of the given equation we can calculate the required value.
Complete step by step answer:
We are given that the equation as
\[{{\left( 64 \right)}^{2}}-{{\left( 36 \right)}^{2}}=20x\]
Now, let us use the formula of difference of squares of two numbers
We know that the formula for difference of square of two numbers that is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
Now, by using this formula to given equation we get
\[\begin{align}
& \Rightarrow \left( 64+36 \right)\left( 64-36 \right)=20x \\
& \Rightarrow 100\times 28=20x \\
\end{align}\]
We know that changing the multiplication from one side of equation to other side it will become division.
Now, by interchanging all the terms except the variable that is taking all terms from RHS to LHS except the variable we get
\[\begin{align}
& \Rightarrow \dfrac{100\times 28}{20}=x \\
& \Rightarrow x=5\times 28 \\
& \Rightarrow x=140 \\
\end{align}\]
Therefore we can conclude that the value of \[x\] is 140.
So, option (d) is the correct answer.
Note:
We can solve this problem without using the formula of algebra.
We are given with the equation that is
\[{{\left( 64 \right)}^{2}}-{{\left( 36 \right)}^{2}}=20x\]
We know that square of a number is multiplying the number with itself that is
\[{{n}^{2}}=n\times n\]
Now, by using the above formula to given equation we get
\[\Rightarrow \left( 64\times 64 \right)-\left( 36\times 36 \right)=20x\]
Now, by applying the required operations in the above equation we get
\[\begin{align}
& \Rightarrow 4096-1296=20x \\
& \Rightarrow 2800=20x \\
\end{align}\]
Now, by cross multiplying the terms that is taking all terms from RHS to LHS except the variable we get
\[\begin{align}
& \Rightarrow \dfrac{2800}{20}=x \\
& \Rightarrow x=\dfrac{280}{2} \\
& \Rightarrow x=140 \\
\end{align}\]
Therefore we can conclude that the value of \[x\] is 140.
So, option (d) is correct answer
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