Answer
397.5k+ views
Hint: In this question, we have an equation. Which have two sides’ left-hand side and right-hand side. First we take the left hand side and solve it.to solve the left hand side we used a formula and formula is given as below.
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
By using the above formula we solve the left hand side, and then we write the left hand side is equal to the right hand side. After that we find the value of\[x\].
Complete step by step answer:
In this question we have given an equation that is,
\[{\left( {x - 3} \right)^2} = 36\]
First, we take the left hand side from the above equation and want to solve it.
Then, the left hand side is.
\[{\left( {x - 3} \right)^2}\]
We know that, \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
Then above the left hand side is written as below.
\[
{\left( {x - 3} \right)^2} \\
= {x^2} + {3^2} - 2\times 3\times x \\
= {x^2} + 9 - 6x \\
\]
Now we write that the right left hand side is equal to the right hand side.
Then,
\[ \Rightarrow {x^2} + 9 - 6x = 36\]
We take \[36\] on the left hand side and on the right hand side.
Then,
Above equation is written as below.
\[ \Rightarrow {x^2} + 9 - 6x - 36 = 0\]
We solve the above equation.
\[{x^2} - 6x - 27 = 0\]
The above equation is a quadratic equation. We solve this equation for the value of\[x\].
Then,
\[
\Rightarrow {x^2} - 6x - 27 = 0 \\
\Rightarrow {x^2} + 3x - 9x - 27 = 0 \\
\]
We take the \[x\]is common in the first two and \[ - 9\]is common in the last two.
Then,
\[
x\left( {x + 3} \right) - 9\left( {x + 3} \right) = 0 \\
\left( {x + 3} \right)\left( {x - 9} \right) = 0 \\
\]
Then,
\[
x + 3 = 0 \\
\therefore x = - 3 \\
\]
And,
\[
x - 9 = 0 \\
\therefore x = 9 \\
\]
Therefore, the values of \[x\] are \[ - 3\] and \[9\].
Note:
In this question, an equation of \[x\] is given, which I want to solve. An equation is defined as it has two things which are equal. And the equation also likes a statement “this equal that”. The equation has two things or two sides, the left side is known as the left hand side and the right side is known as the right hand side. The left-hand side is denoted as “LHS” and the right hand side is denoted as “RHS”.
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
By using the above formula we solve the left hand side, and then we write the left hand side is equal to the right hand side. After that we find the value of\[x\].
Complete step by step answer:
In this question we have given an equation that is,
\[{\left( {x - 3} \right)^2} = 36\]
First, we take the left hand side from the above equation and want to solve it.
Then, the left hand side is.
\[{\left( {x - 3} \right)^2}\]
We know that, \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
Then above the left hand side is written as below.
\[
{\left( {x - 3} \right)^2} \\
= {x^2} + {3^2} - 2\times 3\times x \\
= {x^2} + 9 - 6x \\
\]
Now we write that the right left hand side is equal to the right hand side.
Then,
\[ \Rightarrow {x^2} + 9 - 6x = 36\]
We take \[36\] on the left hand side and on the right hand side.
Then,
Above equation is written as below.
\[ \Rightarrow {x^2} + 9 - 6x - 36 = 0\]
We solve the above equation.
\[{x^2} - 6x - 27 = 0\]
The above equation is a quadratic equation. We solve this equation for the value of\[x\].
Then,
\[
\Rightarrow {x^2} - 6x - 27 = 0 \\
\Rightarrow {x^2} + 3x - 9x - 27 = 0 \\
\]
We take the \[x\]is common in the first two and \[ - 9\]is common in the last two.
Then,
\[
x\left( {x + 3} \right) - 9\left( {x + 3} \right) = 0 \\
\left( {x + 3} \right)\left( {x - 9} \right) = 0 \\
\]
Then,
\[
x + 3 = 0 \\
\therefore x = - 3 \\
\]
And,
\[
x - 9 = 0 \\
\therefore x = 9 \\
\]
Therefore, the values of \[x\] are \[ - 3\] and \[9\].
Note:
In this question, an equation of \[x\] is given, which I want to solve. An equation is defined as it has two things which are equal. And the equation also likes a statement “this equal that”. The equation has two things or two sides, the left side is known as the left hand side and the right side is known as the right hand side. The left-hand side is denoted as “LHS” and the right hand side is denoted as “RHS”.
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