Question

What are the exact solutions of the ${{x}^{2}}-x-4=0$?

Answer 1

Hint: In this problem we need to calculate the solution of the given equation. We can observe that the given equation is a quadratic equation which is in the form of $a{{x}^{2}}+bx+c=0$. So, we will compare the given equation with the equation $a{{x}^{2}}+bx+c=0$ and write the values of $a$, $b$, $c$. We know that the solution for the quadratic equation which is in the form of $a{{x}^{2}}+bx+c=0$ is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We will substitute the values of $a$, $b$, $c$in the above equation and simplify it to get the required solution.

Complete step-by-step solution:
Given equation, ${{x}^{2}}-x-4=0$.
We can observe that the above equation is a quadratic equation which is in the form of $a{{x}^{2}}+bx+c=0$. Comparing the given equation with $a{{x}^{2}}+bx+c=0$, then we will get
$a=1$, $b=-1$, $c=-4$
We know that the solution for the quadratic equation which is in the form of $a{{x}^{2}}+bx+c=0$ is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Substituting the values $a=1$, $b=-1$, $c=-4$ in the above formula to get the solution of the given equation ${{x}^{2}}-x-4=0$, then we will get
$x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -4 \right)}}{2\left( 1 \right)}$
When we multiply a negative sign with a negative sign, we will get positive sign as a result, then the above equation is modified as
$x=\dfrac{1\pm \sqrt{1+16}}{2}$
We know that the value of $1+16$ is given by $17$. Substituting this value in the above equation, then we will get
$x=\dfrac{1\pm \sqrt{17}}{2}$
Hence the solutions of the given equation ${{x}^{2}}-x-4=0$ are $x=\dfrac{1\pm \sqrt{17}}{2}$.

Note: For solving quadratic equations we have a lot of methods like factorization, graphical methods. Both these methods are somewhat lengthy processes and there may be chances for making a lot of mistakes. So, we have not followed those methods. Apart from other methods using quadratic formulas will give you exact solutions of the equations.