Question

How do you solve ${x^2} - 5 = 0$using the quadratic formula?

Answer 1

Hint: As we know that quadratic equations are the equations that are often called second degree. It means that t consists of at least one term that is squared. There are several ways to solve quadratic equations, however using the discriminant is easier, suitable and useful. Suppose the equation will be $a{x^2} + bx + c = 0$, Value of $x$will be $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4\times a\times c} }}{{2a}}$. With this we will get two types of value, one of these will be positive and the other one will be negative. This quadratic formula is also known as Sridharacharya formula.

Complete step-by-step solution:
We have ${x^2} - 5 = 0$. It is in the standard form so here $a = 1,b = 0$and $c = - 5$.
Now by substituting the values into the quadratic formula we get,
$\Rightarrow x = \dfrac{{0 \pm \sqrt {{{(0)}^2} - 4 \times 1 \times ( - 5)} }}{{2 \times 1}}$
$ \Rightarrow x = \dfrac{{0 \pm \sqrt {0 + 20} }}{2}$,
We get $x = \dfrac{{ \pm \sqrt {4 \times 5} }}{2} = \dfrac{{ \pm 2\sqrt 5 }}{2}$,
It gives $x = \pm \sqrt 5 $
Hence the required answer is $ \pm \sqrt 5 $.

Note: We know that the discriminant is the part of the quadratic formula underneath the square root symbol: ${b^2} - 4ac$. It tells us whether there are two solutions, one solution or no solution at all. As in the above solution we got two values one is positive and the other one is negative. It has two real solutions because if the value of discriminant is more than $0$, then the equation has two real solutions. If it is less than $0$, then there are no solutions and if it is equal to $0$then there is only one solution. We should keep in mind that there is one fundamental rule to this quadratic equation which is that the value of $a$, the first constant can never be zero.