Question

How do you solve \[2{{x}^{2}}-4x+7=0\]?

Answer 1

Hint: We will first find the discriminant of the given equation we get, \[D=-40\] which is less than zero. So, the given equation \[2{{x}^{2}}-4x+7=0\] has no real roots or we can say two imaginary roots and which can be found by using the quadratic formula. We will be using iota (\[i\]), which can be written as \[i=\sqrt{(-1)}\]. We get the roots by comparing the given equation with standard form and then substituting the values of a, b and c in the formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step by step solution:
According to the given question, we have to solve for \[x\], so we will try and figure out the type of solution we will get for the given equation.
We will start with finding the discriminant of the given equation,
\[2{{x}^{2}}-4x+7=0\]-----(1)
We know that
\[D={{b}^{2}}-4ac\]
On comparing the discriminant with the equation (1), we get the value of variables as,
\[a=2,b=-4,c=7\]
On substituting these values in the discriminant formula, we get,
\[\Rightarrow D={{(-4)}^{2}}-4(2)(7)\]
\[\Rightarrow D=16-56\]
\[\Rightarrow D=-40<0\]
We have the value of \[D<0\], so the given equation has 2 imaginary roots. We will be using the quadratic formula to solve the given quadratic equation.
We have the quadratic formula as,
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Substituting the values of variables, we have, we get,
\[\Rightarrow x=\dfrac{-(-4)\pm \sqrt{(-56)}}{2(2)}\]
As we can see, we have negative signs in the square root, so in order to solve for \[x\], we will have to deal with this negative sign.
Here, we will use iota (\[i\]), which has the value \[i=\sqrt{(-1)}\], we get,
\[\Rightarrow x=\dfrac{4\pm \sqrt{(7\times 8)}i}{4}\]
\[\Rightarrow x=\dfrac{4\pm 2\sqrt{(7\times 2)}i}{4}\]
Simplifying the terms further, we get,
\[\Rightarrow x=\dfrac{2\pm \sqrt{(14)}i}{2}\]
\[\Rightarrow x=\dfrac{2+\sqrt{14}i}{2},\dfrac{2-\sqrt{14}i}{2}\]

Therefore, we got the two values of \[x=\dfrac{2+\sqrt{14}i}{2},\dfrac{2-\sqrt{14}i}{2}\]

Note: The use of iota (\[i\]) in the above solution is an important one. Without the use of iota, we would not arrive at this final result (even though the roots obtained are imaginary).
We have,
\[{{i}^{2}}=-1\]
\[i=\sqrt{(-1)}\]
By introducing a representation for the square root of minus one, the calculation became convenient and also backed the fact that polynomials having degree ‘n’ have ‘n’ number of roots.