Question

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

Answer 1

In this question we have to find the roots from the above quadratic equation by completing the square. For that we are going to solve simplify the equation. Next, we using PST-Perfect square trinomial. Next, we take the 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 answer:
To find the roots from the given quadratic equation:
From the given $2{x^2} - x - 5 = 0$
First, the factor out the 2 from the first two terms so that the coefficient of ${x^2}$is 1 and we get
$\Rightarrow 2\left( {{x^2} - \dfrac{1}{2}x} \right) - 5 = 0$
Now take note of the $\dfrac{1}{2}$ coefficient of $x$. Divide this $\dfrac{1}{2}$ by 2 it becomes $\dfrac{1}{4}$. Square it, it will become $\dfrac{1}{{16}}$. This $\dfrac{1}{{16}}$ will be added and subtracted to the terms inside the grouping symbol in the above equation.
Let us continue
$\Rightarrow 2\left( {{x^2} - \dfrac{1}{2}x} \right) - 5 = 0$
$\Rightarrow 2\left( {{x^2} - \dfrac{1}{2}x + \dfrac{1}{{16}} - \dfrac{1}{{16}}} \right) - 5 = 0$
Notice ${x^2} - \dfrac{1}{2}x + \dfrac{1}{{16}}$ is a PST-Perfect Square Trinomial. That is ${(a - b)^2} = {a^2} - {b^2} + 2ab$and we get
Put, ${x^2} - \dfrac{1}{2}x + \dfrac{1}{{16}} = {\left( {x - \dfrac{1}{4}} \right)^2}$ in the above equation.
First, we split the parentheses and we get
$\Rightarrow 2\left( {\left( {{x^2} - \dfrac{1}{2}x + \dfrac{1}{{16}}} \right) - \dfrac{1}{{16}}} \right) - 5 = 0$
$\Rightarrow 2\left( {{{\left( {x - \dfrac{1}{4}} \right)}^2} - \dfrac{1}{{16}}} \right) - 5 = 0$
Now, transpose the 5 to the right side then divide both sides by 2 and we get
$\Rightarrow 2\left( {{{\left( {x - \dfrac{1}{4}} \right)}^2} - \dfrac{1}{{16}}} \right) = 5$
$\Rightarrow \dfrac{{{2}\left( {{{\left( {x - \dfrac{1}{4}} \right)}^2} - \dfrac{1}{{16}}} \right)}}{{{2}}} = \dfrac{5}{2}$
$\Rightarrow \left( {{{\left( {x - \dfrac{1}{4}} \right)}^2} - \dfrac{1}{{16}}} \right) = \dfrac{5}{2}$
Then transpose the 1/16 in the right hand side (RHS) and we get
$\Rightarrow {\left( {x - \dfrac{1}{4}} \right)^2} = \dfrac{5}{2} + \dfrac{1}{{16}}$
Next, we take the LCM of right hand side (RHS) and we get
$\Rightarrow {\left( {x - \dfrac{1}{4}} \right)^2} = \dfrac{{(5 \times 8) + 1}}{{16}}$
Simplify the right hand side (RHS) in the above equation.
$\Rightarrow {\left( {x - \dfrac{1}{4}} \right)^2} = \dfrac{{40 + 1}}{{16}}$
On adding the term and we get
$\Rightarrow {\left( {x - \dfrac{1}{4}} \right)^2} = \dfrac{{41}}{{16}}$
Next, we take extract the square root of both sides and we get
$\Rightarrow \sqrt {{{\left( {x - \dfrac{1}{4}} \right)}^2}} = \pm \sqrt {\dfrac{{41}}{{16}}}$
Already, we know that squares and powers are cancelled. So we apply the rule in the above equation and we get
$\Rightarrow x - \dfrac{1}{4} = \pm \dfrac{1}{4}\sqrt {41}$
Here, $\sqrt {\dfrac{1}{{16}}} = \dfrac{1}{4}$
Next, we rearrange the variable and numbers on both sides and we get
$\Rightarrow x = \dfrac{1}{4} \pm \dfrac{1}{4}\sqrt {41}$
On rewriting we get,
$\Rightarrow x = \dfrac{1}{4} \pm \dfrac{{\sqrt {41} }}{4}$
Finally we split the above $x$value we get the required roots.
The roots are
$\Rightarrow {x_1} = \dfrac{1}{4} + \dfrac{{\sqrt {41} }}{4}$
$\Rightarrow {x_2} = \dfrac{1}{4} - \dfrac{{\sqrt {41} }}{4}$
This is the required roots of the given equation.

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 $n$).