
How do you factor \[{x^2} - 6x + 8\]?
Answer
497.4k+ views
Hint: In the given question, we have been asked to factorize the given quadratic equation. To do that, we can easily solve it by using the formula of calculating the value of the variable by using the concept of determinants. We use the formula for solving the value of determinants. Then we just put the value of the determinant into the formula for finding the variable, and that gives us the answer. But it is a point to note that if the value of the determinant is less than 0, i.e., negative, then there is no possible solution for the given equation.
Formula Used:
We are going to use the formula of calculating the value of the variable by using Determinant:
Determinant, \[D = {b^2} - 4ac\]
and, \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
Complete step by step answer:
The given quadratic equation is \[{x^2} - 6x + 8\].
First, we calculate the value of the determinant.
In this question, \[a = 1,{\rm{ }}b = - 6,{\rm{ }}c = 8\].
The formula for finding determinant is:
\[D = {b^2} - 4ac\]
Putting in the values, we get,
$\Rightarrow$ \[D = {\left( { - 6} \right)^2} - 4 \times 1 \times 8 = 36 - 32 = 4 = {\left( 2 \right)^2}\]
Now, we put in the value of \[D\] into the formula for \[x\] and solve,
$\Rightarrow$ \[x = \dfrac{{ - \left( { - 6} \right) \pm 2}}{2} = 3 \pm 1\]
Hence, \[x = 2,4\]
Note: It is not true that whenever there is a given problem, there will be a solution. So, first, we need to check if there are any solutions that can be possible. If there is, then only we proceed to calculate the solution. There is no point in calculating the solution when there is not any. For instance, there is no solution to the point of intersection of two parallel lines. Two parallel lines never intersect, hence, can never have any point of intersection.
Formula Used:
We are going to use the formula of calculating the value of the variable by using Determinant:
Determinant, \[D = {b^2} - 4ac\]
and, \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
Complete step by step answer:
The given quadratic equation is \[{x^2} - 6x + 8\].
First, we calculate the value of the determinant.
In this question, \[a = 1,{\rm{ }}b = - 6,{\rm{ }}c = 8\].
The formula for finding determinant is:
\[D = {b^2} - 4ac\]
Putting in the values, we get,
$\Rightarrow$ \[D = {\left( { - 6} \right)^2} - 4 \times 1 \times 8 = 36 - 32 = 4 = {\left( 2 \right)^2}\]
Now, we put in the value of \[D\] into the formula for \[x\] and solve,
$\Rightarrow$ \[x = \dfrac{{ - \left( { - 6} \right) \pm 2}}{2} = 3 \pm 1\]
Hence, \[x = 2,4\]
Note: It is not true that whenever there is a given problem, there will be a solution. So, first, we need to check if there are any solutions that can be possible. If there is, then only we proceed to calculate the solution. There is no point in calculating the solution when there is not any. For instance, there is no solution to the point of intersection of two parallel lines. Two parallel lines never intersect, hence, can never have any point of intersection.
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