Question

How do you solve ${x^2} + 3x - 5 = 0$ by completing the square?

Answer 1

Hint: This equation is the quadratic equation. The general form of the quadratic equation is $a{x^2} + bx + c = 0$. Where ‘a’ is the coefficient of${x^2}$, ‘b’ is the coefficient of x and ‘c’ is the constant term. In this question, we want to solve the quadratic equation by completing the square method. In this method, we first make the leading coefficient of an equation one, if it is not one by dividing the whole equation with the leading coefficient and then add ${\left( {\dfrac{{ab}}{2}} \right)^2}$ on both sides of the above equation. Then solve further by taking square-root to get roots of the given equation.

Complete step by step solution:
In this question, we want to solve the quadratic equation by completing the square method.
The given equation is,
$ \Rightarrow {x^2} + 3x - 5 = 0$
There are different steps to find the roots.
The first step is to shift the constant term to the right-hand side.
Therefore, let us add 5 on both sides.
$ \Rightarrow {x^2} + 3x - 5 + 5 = 0 + 5$
That is equal to,
$ \Rightarrow {x^2} + 3x = 5$
The second step is to remove the coefficient of ${x^2}$ if it is there.
The third step is to add ${\left( {\dfrac{{ab}}{2}} \right)^2}$ on both sides of the above equation. Here, ‘a’ is 1 and ‘b’ is 3.
So, add ${\left( {\dfrac{{1 \times 3}}{2}} \right)^2} = {\left( {\dfrac{3}{2}} \right)^2}$
 $ \Rightarrow {x^2} + 3x + {\left( {\dfrac{3}{2}} \right)^2} = 5 + {\left( {\dfrac{3}{2}} \right)^2}$
So,
$ \Rightarrow {\left( x \right)^2} + 2\left( 1 \right)\left( {\dfrac{3}{2}} \right)x + {\left( {\dfrac{3}{2}} \right)^2} = 5 + \dfrac{9}{4}$
The left-hand side is in the form of ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. Also, take LCM on the right-hand side.
Here, the value of ‘a’ is x and the value of ‘b’ is $\dfrac{3}{2}$ .
$ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{{20 + 9}}{4}$
That is equal to,
$ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{{29}}{4}$
Let us take square roots on both sides.
$ \Rightarrow \sqrt {{{\left( {x + \dfrac{3}{2}} \right)}^2}} = \pm \sqrt {\dfrac{{29}}{4}} $
Therefore,
$ \Rightarrow x + \dfrac{3}{2} = \pm \dfrac{{\sqrt {29} }}{2}$
Let us subtract $\dfrac{3}{2}$ on both sides.
$ \Rightarrow x + \dfrac{3}{2} - \dfrac{3}{2} = \pm \dfrac{{\sqrt {29} }}{2} - \dfrac{3}{2}$
So,
$ \Rightarrow x = \pm \dfrac{{\sqrt {29} }}{2} - \dfrac{3}{2}$

Hence, the roots of the given quadratic equation is $\dfrac{{\sqrt {29} }}{2} - \dfrac{3}{2}$ and $ - \dfrac{{\sqrt {29} }}{2} - \dfrac{3}{2}$.

Note:
Here is a list of methods to solve quadratic equations:
1) Factorization
2) Completing the square
3) Using graph
4) Quadratic formula