Question

How do you solve ${{x}^{2}}-2x+1=4$ by completing the square?

Answer 1

Hint: To solve the given equation by the method of completing the square, we will have to convert the given equation in the form of a square so that our process can be on the right track of the solution method that is completing the square method. So, for making the complete square, we will use convert the equation in the form of ${{a}^{2}}+2ab+{{b}^{2}}$ or ${{a}^{2}}-2ab+{{b}^{2}}$ that will be equal to ${{\left( a+b \right)}^{2}}$ or ${{\left( a-b \right)}^{2}}$ respectively. After that we will simplify the equation to get the value of the variable.

Complete step by step solution:
Since, we have the question in the form of quadratic equation as:
$\Rightarrow {{x}^{2}}-2x+1=4$
Here, we need to convert the above equation in the form of ${{a}^{2}}+2ab+{{b}^{2}}$ or ${{a}^{2}}-2ab+{{b}^{2}}$ . So, we have one variable in the form of a square that is ${{x}^{2}}$ and the other one is $1$ . Here, we can write $1$ in the form of square as ${{1}^{2}}$ because the square of $1$ is always $1$ and we need a term in the form of $2ab$, so we already have $2x$ that can be written as $2\times x\times 1$ . Thus the above equation will be as:
$\Rightarrow {{x}^{2}}-2\times x\times 1+{{1}^{2}}=4$
Now, we got the equation in the form of ${{a}^{2}}-2ab+{{b}^{2}}$ that is equal to ${{\left( a-b \right)}^{2}}$ . So, we can write the above equation in the form of ${{\left( x-1 \right)}^{2}}$ and here, we can write $4$as a square of $2$ as ${{2}^{2}}$ because ${{2}^{2}}$ is equal to $4$ . Therefore, the above equation will be:
$\Rightarrow {{\left( x-1 \right)}^{2}}={{2}^{2}}$
Now, we will take the square root both sides of the above equation as:
$\Rightarrow x-1=\pm 2$
Since, squares of $+2$ and $-2$ are $4$ that is the same value. So, we will write both the values as a square root of $4$ .
Now, we will solve the equation in both cases as:

Case I:
$\Rightarrow x-1=+2$
Now, we will do necessary calculation for the above equation as:
$\Rightarrow x=2+1$
$\Rightarrow x=3$

Now, Case II:
$\Rightarrow x-1=-2$
$\Rightarrow x=-2+1$
After calculating the above equation as:
$\Rightarrow x=-1$

Hence, from the both cases, we had two values of $x$ that are $3$ and $-1$ .

Note: Here, we will verify the solution if it is correct or not by applying the both values of $x$ in the given equation as:
Here, we will apply $x=3$
$\Rightarrow {{3}^{2}}-2\times 3+1=4$
Now, we will solve the left hand side of the equation as:
$\Rightarrow 9-6+1=4$
Here, we will add both the positive numbers as:
$\Rightarrow 10-6=4$
Now, we will use subtraction as:
$\Rightarrow 4=4$
Thus, we got that \[\text{LHS}=\text{RHS}\] .

Now, we will apply $x=-1$ in the given equation as:
$\Rightarrow {{\left( -1 \right)}^{2}}-2\times \left( -1 \right)+1=4$
After doing required calculation we will get:
$\Rightarrow 1+2+1=4$
$\Rightarrow 4=4$
Again we got \[\text{LHS}=\text{RHS}\] .
Hence, the solution is correct.