Question

How do you solve $2{x^2} - 5x - 12 = 0$ by completing the square?

Answer 1

Hint: In this question we have to find the roots from the above quadratic equation by completing the square. For that we are going to simplify the equation. Next, we extract the square root of both sides and then simplify to arrive at our final answer. And also we are going to add and subtraction in complete step by step solution.

Complete step by step solution:
To find the roots from the given quadratic equation:
$ \Rightarrow 2{x^2} - 5x - 12 = 0$
Divide both sides of the equation by 2 to have 1 as the coefficient of the first term :
$ \Rightarrow {x^2} - \dfrac{5}{2}x - 12 = 0$
$ \Rightarrow {x^2} - \left( {\dfrac{5}{2}} \right)x - 6{\text{ }} = {\text{ }}0$

Add 6 to both side of the equation:
$ \Rightarrow {x^2} - \left( {\dfrac{5}{2}} \right)x = 6$
Now the clever bit: Take the coefficient of x , which is\[\dfrac{5}{2}\] , divide by two, giving \[\dfrac{5}{4}\] , and finally square it giving \[\dfrac{{25}}{{16}}\]
Add \[\dfrac{{25}}{{16}}\] to both sides of the equation :
On the right hand side we have:
\[6\;\; + \;\;\dfrac{{25}}{{16}}\] or, \[\left( {\dfrac{6}{1}} \right) + \left( {\dfrac{{25}}{{16}}} \right)\;\]
The common denominator of the two fractions is 16 Adding \[\left( {\dfrac{{96}}{{16}}} \right) + \left( {\dfrac{{25}}{{16}}} \right)\;\]gives \[\dfrac{{121}}{{16\;}}\]
So adding to both sides we finally get :
\[ \Rightarrow {x^2} - \left( {\dfrac{5}{2}} \right)x + \left( {\dfrac{{25}}{{16}}} \right){\text{ }} = {\text{ }}\dfrac{{121}}{{16}}\]
Adding \[\dfrac{{25}}{{16}}\] has completed the left hand side into a perfect square:
\[ \Rightarrow \left( {x - \left( {\dfrac{5}{4}} \right)} \right)\; \times \left( {x - \left( {\dfrac{5}{4}} \right)} \right)\;\;\]
\[ \Rightarrow {x^2} - \left( {\dfrac{5}{2}} \right)x + \left( {\dfrac{{25}}{{16}}} \right)\]
\[ \Rightarrow {\left( {x - \left( {\dfrac{5}{4}} \right)} \right)^2}\]
Things which are equal to the same thing are also equal to one another.
Since\[ \Rightarrow {x^2} - \left( {\dfrac{5}{2}} \right)x + \left( {\dfrac{{25}}{{16}}} \right){\text{ }} = {\text{ }}\dfrac{{121}}{{16}}\] and
\[ \Rightarrow {x^2} - \left( {\dfrac{5}{2}} \right)x + \left( {\dfrac{{25}}{{16}}} \right) = {\left( {x - \left( {\dfrac{5}{4}} \right)} \right)^2}\]
Then, according to the law of transitivity,
\[ \Rightarrow {\left( {x - \left( {\dfrac{5}{4}} \right)} \right)^2} = \dfrac{{121}}{{16}}\]
The square root principle says that When two things are equal, their square roots are equal.
Note that the square root of
\[ \Rightarrow \;{\left( {x - \left( {\dfrac{5}{4}} \right)} \right)^2}\] is
\[ \Rightarrow {\left( {x - \left( {\dfrac{5}{4}} \right)} \right)^{2/2}}\;\]
\[ \Rightarrow {\left( {x - \left( {\dfrac{5}{4}} \right)} \right)^1}\]
\[ \Rightarrow x - \left( {\dfrac{5}{4}} \right)\]
Now, applying square root principle to we get:
\[ \Rightarrow \left( {x - \left( {\dfrac{5}{4}} \right)} \right) = \sqrt {\dfrac{{121}}{{16}}} \]
Add $\dfrac{5}{4}$ to both sides to obtain:
\[ \Rightarrow x = \left( {\dfrac{5}{4}} \right) \pm \sqrt {\dfrac{{121}}{{16}}} \]
Since a square root has two values, one positive and the other negative
$ \Rightarrow {x^2} - \left( {\dfrac{5}{2}} \right)x - 6{\text{ }} = {\text{ }}0$ has two solutions:
\[\; \Rightarrow \;x{\text{ }} = {\text{ }}\dfrac{5}{4}{\text{ }} + {\text{ }}\sqrt {\dfrac{{121}}{{16}}} \]or
\[\; \Rightarrow \;x{\text{ }} = {\text{ }}\dfrac{5}{4}{\text{ - }}\sqrt {\dfrac{{121}}{{16}}} \]
Note that \[\;\sqrt {\dfrac{{121}}{{16}}} \] can be written as\[\;\dfrac{{\sqrt {121} }}{{\sqrt {16} }}\]which is $\dfrac{{11}}{4}$

Note: The students only remember completing the square. That is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.
The square of an expression of the form $x + n$ or $x - n$ is :
\[{(x + n)^2} = {x^2} + 2nx + {n^2}\]or \[{(x - n)^2} = {x^2} - 2nx + {n^2}\].
Notice the sign on the middle term matches the sign in the middle of the binomial on the left and the last term is positive in both.
Also notice that if we allow $n$ to be negative, we only need to write and think about\[{(x + n)^2} = {x^2} + 2nx + {n^2}\] (The sign in the middle will match the sign of $n$).
In relation to quadratic equations, imaginary numbers (and complex numbers) occur when the value under the radical portion of the quadratic formula is negative. When this occurs, the equation has no roots (zeros) in the set of real numbers. The roots belong to the set of complex numbers, and will be called "complex roots" (or "imaginary roots"). These complex roots will be expressed in the form $a + ib$.