Question

How do you solve $3{{x}^{2}}+5x-2=0$ ?

Answer 1

Hint: Quadratic equations of this type can be solved by completing the square method. We will add some numbers on both the sides and rearrange some terms to have a complete square in the form of ${{\left( x+a \right)}^{2}}$ . Now, the solution of the equation can be found by further simplifying the equation.

Complete step-by-step solution:
The given equation is
$3{{x}^{2}}+5x-2=0$
We first start by adding $2$ to both sides of the equation
$\Rightarrow 3{{x}^{2}}+5x=2$
Dividing both sides by $3$ we get
$\Rightarrow {{x}^{2}}+\dfrac{5}{3}x=\dfrac{2}{3}$
We rewrite the equation as
$\Rightarrow {{x}^{2}}+2\cdot \dfrac{5}{3\times 2}x=\dfrac{2}{3}$
$\Rightarrow {{x}^{2}}+2\cdot \dfrac{5}{6}x=\dfrac{2}{3}....\text{expression}1$
To have a complete square in the left-hand side, let’s take the square ${{\left( x+a \right)}^{2}}$ for comparison.
We know, ${{\left( x+a \right)}^{2}}={{x}^{2}}+2\cdot a\cdot x+{{a}^{2}}....\text{expression2}$
As, the first two terms of the expression $\left( {{x}^{2}}+2\cdot a\cdot x+{{a}^{2}} \right)$ are ${{x}^{2}}$ and $2x$ , we compare the left hand side of$\text{expression}1$ with the right-hand side of $\text{expression2}$ and we get $a=\dfrac{5}{6}$ .
Hence, to get the square term ${{\left( x+\dfrac{5}{6} \right)}^{2}}$ we add $\dfrac{25}{36}$ to the both sides of $\text{expression}1$
$\Rightarrow {{x}^{2}}+2\cdot \dfrac{5}{6}x+\dfrac{25}{36}=\dfrac{2}{3}+\dfrac{25}{36}$
$\Rightarrow {{x}^{2}}+2\cdot \dfrac{5}{6}x+\dfrac{25}{36}=\dfrac{49}{36}$
The above equation can be also written as
$\Rightarrow {{\left( x+\dfrac{5}{6} \right)}^{2}}=\dfrac{49}{36}$
Now taking square root on both the sides of the equation and also keeping both the values after doing the square root we get
$\Rightarrow x+\dfrac{5}{6}=\pm \dfrac{7}{6}$
Further simplifying we get
$\Rightarrow x=\dfrac{1}{3}$ and $x=-2$
Therefore, the solution of the equation $3{{x}^{2}}+5x-2=0$ are $x=\dfrac{1}{3}$ and $x=-2$.

Note: We have to keep in mind while simplifying at the last part of the solution we have to take both the signs of $\dfrac{7}{6}$ , otherwise inaccuracy in solution will occur. The given problem can also be solved by other methods, such as by factoring the left-hand side of the equation and equating both the factors individually to zero. Also, it can be done by using the Sridhar Acharya formula for finding the solution, which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .