Question

How do you solve $4{{x}^{2}}-2x-5=0$ by completing the square?

Answer 1

A quadratic equation can be solved easily by completing the square after going through some rearrangements of the terms. We must have a complete square in the form of ${{\left( x-a \right)}^{2}}$ by adding some numbers on both the side and convert the left-hand side into a complete square. Now, further simplifying the equation we get the value of $x$ .

Complete step-by-step solution:
The equation we have is
$4{{x}^{2}}-2x-5=0$
Adding $5$ to both the sides of the above equation we get
$\Rightarrow 4{{x}^{2}}-2x=5$
Dividing both the sides of the above equation we get
$\Rightarrow {{x}^{2}}-\dfrac{1}{2}x=\dfrac{5}{4}$
The above expression can also be written as
$\Rightarrow {{x}^{2}}-2\cdot \dfrac{1}{4}\cdot x=\dfrac{5}{4}....\text{expression}1$
To have a complete square in the left-hand side of the equation let’s start comparing the given equation with the square ${{\left( x-a \right)}^{2}}$
We know, ${{\left( x-a \right)}^{2}}={{x}^{2}}-2\cdot a\cdot x+{{a}^{2}}....\text{expression2}$
We can see that the first two terms of the expression $\left( {{x}^{2}}-2\cdot a\cdot x+{{a}^{2}} \right)$ are ${{x}^{2}}$ and $2\cdot a\cdot x$ , we compare $\text{expression}1$ with the first two terms in the right-hand side of $\text{expression2}$ and we get $a=\dfrac{1}{4}$ .
Hence, to get the square term ${{\left( x-\dfrac{1}{4} \right)}^{2}}$ we add $\dfrac{1}{16}$ to the both sides of $\text{expression}1$
$\Rightarrow {{x}^{2}}-2\cdot \dfrac{1}{4}\cdot x+\dfrac{1}{16}=\dfrac{5}{4}\text{+}\dfrac{1}{16}$
After further simplification of the above expression, we get
$\Rightarrow {{x}^{2}}-2\cdot \dfrac{1}{4}\cdot x+\dfrac{1}{16}=\dfrac{21}{16}$
The left-hand side of the above equation can be also written as
$\Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}=\dfrac{21}{16}$
Now taking square root on both the sides of the above equation and we keep both the values after doing the square root as shown below
$\Rightarrow \left( x-\dfrac{1}{4} \right)=\pm \dfrac{\sqrt{21}}{4}$
Further doing some simplification we get
$\Rightarrow x=\pm \dfrac{\sqrt{21}}{4}+\dfrac{1}{4}$
$\Rightarrow x=\dfrac{1\pm \sqrt{21}}{4}$

Therefore, the solution of the given equation is $x=\dfrac{1+\sqrt{21}}{4}$ and $x=\dfrac{1-\sqrt{21}}{4}$.

Note: To balance the both sides of an equation properly we must be careful while taking the terms to the other side of the equation. Also, we must keep in mind that after doing the square root on both the sides of the equation we must take both the positive as well as negative signs to get both the solutions.