How do you factor \[{x^2} - x + 3 = 0\] using the quadratic formula?
Answer
560.7k+ 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 \[{x^2} - x + 3 = 0\].
First, we calculate the value of the determinant.
In this question, \[a = 1,{\rm{ }}b = - 1,{\rm{ }}c = 3\].
The formula for finding determinant is:
\[D = {b^2} - 4ac\]
Putting in the values, we get,
\[D = {\left( { - 1} \right)^2} - 4 \times 1 \times 3 = 1 - 12 = - 11 < 0\]
We got the value of determinant which is less than zero.
Hence, the given equation has no real solutions.
Note:In the given question, we had to factorize the given polynomial. The biggest mistake that some students make is to start factoring the polynomial by splitting the middle without checking the determinant and end up wasting a lot of time. If the calculated value of determinant is negative, there is no possibility of a real solution. So, before solving by splitting the middle term, we need to check the value of determinant and then solve.
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 \[{x^2} - x + 3 = 0\].
First, we calculate the value of the determinant.
In this question, \[a = 1,{\rm{ }}b = - 1,{\rm{ }}c = 3\].
The formula for finding determinant is:
\[D = {b^2} - 4ac\]
Putting in the values, we get,
\[D = {\left( { - 1} \right)^2} - 4 \times 1 \times 3 = 1 - 12 = - 11 < 0\]
We got the value of determinant which is less than zero.
Hence, the given equation has no real solutions.
Note:In the given question, we had to factorize the given polynomial. The biggest mistake that some students make is to start factoring the polynomial by splitting the middle without checking the determinant and end up wasting a lot of time. If the calculated value of determinant is negative, there is no possibility of a real solution. So, before solving by splitting the middle term, we need to check the value of determinant and then solve.
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