Question

How do you solve using the quadratic formula: \[{{x}^{2}}+x+4=0\]?

Answer 1

Hint: Suppose that we have a quadratic equation in the form \[a{{x}^{2}}+bx+c=0\] where a, b and c is any number. Using quadratic formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], we can calculate the roots of the given equation. The behaviour of any quadratic equation \[a{{x}^{2}}+bx+c=0\] can be predicted by its discriminant given by \[{{b}^{2}}-4ac\].

Complete step by step answer:
We know that the sum of roots of this quadratic equation, that is \[{{x}_{1}}+{{x}_{2}}\] equals to negative of the ratio of x- coefficient to \[{{x}^{2}}\] - coefficient \[\Rightarrow \dfrac{-b}{a}\], which we write as \[{{x}_{1}}+{{x}_{2}}=\dfrac{-b}{a}\]. Also, the product of the roots of this quadratic equation, that is \[{{x}_{1}}\times {{x}_{2}}\] equals to the ratio of constant term to \[{{x}^{2}}\] - coefficient \[\Rightarrow \dfrac{c}{a}\].
That is, we can write it as \[{{x}_{1}}\times {{x}_{2}}=\dfrac{c}{a}\].
Using algebraic formula \[{{({{x}_{1}}-{{x}_{2}})}^{2}}={{({{x}_{1}}+{{x}_{2}})}^{2}}-4({{x}_{1}}\times {{x}_{2}})\], we get \[({{x}_{1}}-{{x}_{2}})=\dfrac{\sqrt{{{b}^{2}}-4ac}}{a}\].
From the above equation, we have the quadratic formula which gives the information about the roots of the quadratic equation: \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
In the given question, we are given the quadratic equation \[{{x}^{2}}+x+4=0\]. Here, 1, 1 and 4 are the \[{{x}^{2}}\] - coefficient, x - coefficient and constant term respectively.
Compared with the general quadratic equation \[a{{x}^{2}}+bx+c=0\], we can write a = 1, b = 1 and c = 4. Then, \[{{x}_{1}}+{{x}_{2}}=\dfrac{-1}{1}\] which is simply -1 and \[{{x}_{1}}\times {{x}_{2}}=\dfrac{4}{1}\] which is simply 4.
On plugging the values of coefficient of \[{{x}^{2}}\] (a), coefficient of x (b) and the constant term (c) into the quadratic formula: \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], we get
\[\Rightarrow x=\dfrac{-(1)\pm \sqrt{{{(1)}^{2}}-4(1)(4)}}{2(1)}\]
\[\Rightarrow x=\dfrac{-1\pm \sqrt{1-16}}{2}\]
\[\Rightarrow x=\dfrac{-1\pm \sqrt{-15}}{2}\]
Here, we have (-15) under square root. This gives a complex number equal to \[\sqrt{15}i\] since \[\sqrt{-1}=i\]. Then,
\[\Rightarrow x=\dfrac{-1\pm \sqrt{15}i}{2}\]
\[\therefore x=\dfrac{-1+\sqrt{15}i}{2}\] and \[x=\dfrac{-1-\sqrt{15}i}{2}\] are complex roots of the given quadratic equation \[{{x}^{2}}+x+4=0\].
Actually, if we consider only real roots then we can say that this quadratic equation doesn’t have roots. This means that there is no solution for this quadratic equation.

Note: We can also solve quadratic equations by factoring methods in which we try to obtain a common factor by splitting up the x term. We can understand the behaviour of roots of any quadratic equation from the value of discriminant D = \[{{b}^{2}}-4ac\] where a, b and c are \[{{x}^{2}}\] - coefficient, x - coefficient and constant term respectively. Only in case of negative value of discriminant, we have no real solution for the quadratic equation.