Question

How do you solve $2{{x}^{2}}+3x=20$ by completing the square?

Answer 1

To solve the equation $a{{x}^{2}}+bx+c=0$ by completing the square method, write the equation such that c (constant part) is on the right side of the equality sign i.e., $a{{x}^{2}}+bx=c$. First, we will convert ‘a’ into 1. Then, add the square of the half of the coefficient of x i.e., ${{\left( \dfrac{b}{2a} \right)}^{2}}$ on both sides of the equation. After that, the left side will become ${{\left( x+\dfrac{b}{2a} \right)}^{2}}$and then take the square root of both sides. After taking square roots, take both possibilities of plus and minus and then just solve the expressions for x and there will be two values of x.

Complete step-by-step solution:
It is given in the question that our equation is $2{{x}^{2}}+3x=20$ and we have been asked to find the value of x using completing the square method.
To solve the equation $a{{x}^{2}}+bx+c=0$ by completing the square method, we will first write the equation such that c (constant part) is on the right side of the equality sign i.e., $a{{x}^{2}}+bx=c$. And we already have the same type of equation.
Therefore, on comparing, we get
a = 2; b = 3; c = 20;
Here, our constant term is already on the right side but the coefficient of ${{x}^{2}}$ is 2. So, to convert the coefficient of ${{x}^{2}}$ into 1, we have to divide the equation by 2 both sides.
Now, we will divide the equation by 2.
$\Rightarrow \dfrac{\left( 2{{x}^{2}}+3x \right)}{2}=\dfrac{20}{2}$
On further simplifications, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x=10$
Now, we will add ${{\left( \dfrac{3}{2\times 2} \right)}^{2}}={{\left( \dfrac{3}{4} \right)}^{2}}$ on both sides of the equation, we get
$\Rightarrow {{x}^{2}}+\dfrac{3}{2}x+{{\left( \dfrac{3}{4} \right)}^{2}}=10+{{\left( \dfrac{3}{4} \right)}^{2}}$
And, we can further write it as
$\Rightarrow {{x}^{2}}+2\times \dfrac{3}{4}x+{{\left( \dfrac{3}{4} \right)}^{2}}=10+{{\left( \dfrac{3}{4} \right)}^{2}}$
And we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, therefore, we get
$\Rightarrow {{\left( x+\dfrac{3}{4} \right)}^{2}}=10+{{\left( \dfrac{3}{4} \right)}^{2}}$
And on further simplifications, we get
$\Rightarrow {{\left( x+\dfrac{3}{4} \right)}^{2}}=10+\dfrac{9}{16}$
$\Rightarrow {{\left( x+\dfrac{3}{4} \right)}^{2}}=\dfrac{160+9}{16}$
$\Rightarrow {{\left( x+\dfrac{3}{4} \right)}^{2}}=\dfrac{169}{16}$
Now, we will take the square root on both sides of the equation. Therefore, we get
$\Rightarrow \sqrt{{{\left( x+\dfrac{3}{4} \right)}^{2}}}=\pm \sqrt{\dfrac{169}{16}}$
And we know that $\dfrac{169}{16}={{\left( \dfrac{13}{4} \right)}^{2}}$
$\Rightarrow \sqrt{{{\left( x+\dfrac{3}{4} \right)}^{2}}}=\pm \sqrt{{{\left( \dfrac{13}{4} \right)}^{2}}}$
And we can also write it as
$\Rightarrow x+\dfrac{3}{4}=\pm \dfrac{13}{4}$
Now, case 1 will be with the plus sign.
Case 1:
$x+\dfrac{3}{4}=\dfrac{13}{4}$
$\Rightarrow x=\dfrac{13}{4}-\dfrac{3}{4}$
And on further simplification, we get
$\Rightarrow x=\dfrac{10}{4}$
And it can also be written as
$\Rightarrow x=\dfrac{5}{2}$
And case 2 will be with the minus sign.
Case 2:
$x+\dfrac{3}{4}=-\dfrac{13}{4}$
$\Rightarrow x=-\dfrac{13}{4}-\dfrac{3}{4}$
And on further simplification, we get
$\Rightarrow x=-\dfrac{16}{4}$
And it can also be written as
$\Rightarrow x=-4$
So, we have found the solution of the equation $2{{x}^{2}}+3x=20$ by completing the square method as $x=\dfrac{5}{2}$, and $x=-4$.
Hence, the solutions of the given equation $2{{x}^{2}}+3x=20$ by completing the square method are $x=\dfrac{5}{2},-4$.

Note: Whenever we get this type of problem, try to make the left side a whole square. We can also do this problem by hit and trial method and also with the graph method. We should not make mistakes in the calculation part and always take care of signs while moving terms from left to right and vice-versa. After taking the square root, there are two possibilities i.e., plus and minus.