Question

Which of the following must be added to \[{x^2} - 6x + 5\] to make it a perfect square?
(A) $3$
(B) $4$
(C) $5$
(D) $6$

Answer 1

Hint: In this question, we will use an identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$. Then we will convert the given equation in the format of the above identity and add the desired number to it to make the perfect square. Here we have to add a positive number from the given options.

Complete step by step answer:
In the above question, we have to add a number to \[{x^2} - 6x + 5\] to make it a perfect square from the given options.
Here we need to use the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ to make it a perfect square.
We have,
\[{x^2} - 6x + 5\]
We can also write it as
${x^2} - 2 \times 3 \times x + 5$
On comparing it with ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$, we get $a = x\,\,and\,\,b = 3$
Therefore, ${b^2} = 9$
So, to get $9$ in the above equation we have to add $4$ to it so that we can form a perfect square.
Therefore, on adding
${x^2} - 2 \times 3 \times x + 5 + 4$
$ \Rightarrow {x^2} - 2 \times 3 \times x + 9$
We can also write it as
$ \Rightarrow {x^2} - 2 \times 3 \times x + {\left( 3 \right)^2}$
Therefore, on using the identity we get
$ \Rightarrow {\left( {x - 3} \right)^2}$
Hence, the correct option is (B).

Note:
If a quadratic equation forms a perfect square, then the roots of the equation are real and equal. We know that a quadratic equation has two roots, they can be either equal or not. But in the case of a perfect square both the roots are equal. Also, the value of discriminant in this case is zero.
So, we easily find the root of the equation using the quadratic formula. The expression of discriminant is $D = {b^2} - 4ac$.