Question

How do you solve \[{x^2} + 3x - 9 = 0\] by completing the square?

Answer 1

Hint: We need to add a constant term of equal quantity to the both sides to find the square term. On doing some simplification we get the required answer.

Formula used: We have to apply following algebraic formula:
\[{(a + b)^2} = ({a^2} + 2ab + {b^2}).\]

Complete step-by-step solution:
The given expression is as following:
\[{x^2} + 3x - 9 = 0\].
So, if we tally the L.H.S in the above equation with the formula of \[{(a + b)^2}\], we can say that the term \[2ab\] should be equivalent to the term \[3x\].
So, if we expand the term \[3x\], we can re-write it as \[\left( {2 \times x \times \dfrac{3}{2}} \right)\].
Therefore, we can say that the term \[a\]and \[b\]in the formula are equivalent to \[x\] and \[\dfrac{3}{2}\].
So, we need to add a term in both of the sides of the equation that is squared of \[\dfrac{3}{2}\].
So, we have to add \[{\left( {\dfrac{3}{2}} \right)^2} = \dfrac{9}{4}\]on the both the sides of the given equation.
So, we can re-write the equation as following:
\[ \Rightarrow {x^2} + 3x + \dfrac{9}{4} - 9 = \dfrac{9}{4}.\]
Now, take the constant term, \[ - 9\] in the R.H.S, we get the following equation:
\[ \Rightarrow {x^2} + 3x + \dfrac{9}{4} = \dfrac{9}{4} + 9.\]
Now, if we write the terms in the L.H.S as in the expanded form, we get:
\[ \Rightarrow {x^2} + 2 \times x \times \dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} = \dfrac{{9 + 36}}{4} = \dfrac{{45}}{4}..............(1)\]
So, by comparing the L.H.S in the above equation with the formula: \[{(a + b)^2} = ({a^2} + 2ab + {b^2})\], we re-write L.H.S as following:
\[ \Rightarrow {x^2} + 2 \times x \times \dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} = {\left( {x + \dfrac{3}{2}} \right)^2}\].
So, we can re-write the equation\[(1)\]as following:
\[ \Rightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{{45}}{4}.\].
Now, if we take the square root on both the sides of the above equation, we can write the following equation:
\[\sqrt {{{(x + \dfrac{3}{2})}^2}} = \pm \sqrt {\dfrac{{45}}{4}} .\]. (We use \[ \pm \] sign as the square root can give us either positive or negative value)
After solving the above equation, we get: \[\left( {x + \dfrac{3}{2}} \right) = \pm \dfrac{{3\sqrt 5 }}{2}.\].
Now, take the constant term to the R.H.S from the L.H.S, we get:\[x = - \dfrac{3}{2} \pm \dfrac{{3\sqrt 5 }}{2} = - \dfrac{3}{2}\left( {1 \pm \sqrt 5 } \right).\]

\[\therefore \]The given equation have two different roots of \[ - \dfrac{3}{2}\left( {1 + \sqrt 5 } \right)\] and \[ - \dfrac{3}{2}\left( {1 - \sqrt 5 } \right)\].

Note: Points to remember:
If we multiply one term by an even number of times, it will always give us a positive sign before it.
But if we multiply one term an odd number of times, it can give either a positive or negative sign depending on the sign it carries.