Question

Solve the given quadratic equation for x: $x\left( 2x+5 \right)=3$

Answer 1

- Hint: First of all open the bracket of the left hand side of the equation and then we get the quadratic equation in x as $2{{x}^{2}}+5x-3=0$.We can solve the above quadratic equation by factorization method. In the factorization method, we will split the term 5x in such a way that the split terms of 5x on multiplication will yield $2\times 6$.

Complete step-by-step solution -

The equation given in the question of which we have to find solutions is:
$x\left( 2x+5 \right)=3$
Opening the bracket of L.H.S of the above equation we get,
$2{{x}^{2}}+5x=3$
Rearranging the above equation in the form of $f\left( x \right)=0$ we get,
$2{{x}^{2}}+5x-3=0$
We are going to split 5 in 5x in such a way that the terms in the split on multiplication yield $2\times 3$.
Now, the factors of $2\times 3$ are $3\times 2\times 1$.
From the factors of $2\times 3$ we can write $2\times 3$ as $1\times 6$.
If we subtract 1 from 6, we will get 5 which is the coefficient of x in the given quadratic equation.
Splitting $5x$ as:
$5x=6x-x$
Substituting the above value of $5x$ in the given quadratic equation we get,
$\begin{align}
  & 2{{x}^{2}}+5x-3=0 \\
 & \Rightarrow 2{{x}^{2}}+6x-x-3=0 \\
\end{align}$
$\begin{align}
  & \Rightarrow 2x\left( x+3 \right)-1\left( x+3 \right)=0 \\
 & \Rightarrow \left( 2x-1 \right)\left( x+3 \right)=0 \\
\end{align}$
Now, equating $2x-1=0$ and $x+3=0$ we get:
$\begin{align}
  & 2x-1=0;x+3=0 \\
 & \Rightarrow x=\dfrac{1}{2},-3 \\
\end{align}$
Hence, the solution of the given quadratic equation is $x=\dfrac{1}{2},-3$.

Note: The alternative way of solving the above quadratic equation as follows:
$2{{x}^{2}}+5x-3=0$
 We are solving the roots of the equation by discriminant formula.
Discriminant of a quadratic equation is denoted by D.
For the quadratic equation $a{{x}^{2}}+bx+c=0$ the value of D is:
$D={{b}^{2}}-4ac$
Comparing this value of D with the given quadratic equation $2{{x}^{2}}+5x-3=0$ we get,
$\begin{align}
  & D={{\left( 5 \right)}^{2}}-4\left( 2 \right)\left( -3 \right) \\
 & \Rightarrow D=25+24 \\
 & \Rightarrow D=49 \\
\end{align}$
The discriminant formula for finding the roots of quadratic equation $a{{x}^{2}}+bx+c=0$ is:
$x=\dfrac{-b\pm \sqrt{D}}{2a}$
Comparing the above value of x with the given quadratic equation $2{{x}^{2}}+5x-3=0$ we get,
$\begin{align}
  & x=\dfrac{-5\pm \sqrt{49}}{4} \\
 & \Rightarrow x=\dfrac{-5\pm 7}{4} \\
\end{align}$
Taking plus sign we get the value of x as:
$\begin{align}
  & x=\dfrac{-5+7}{4} \\
 & \Rightarrow x=\dfrac{2}{4}=\dfrac{1}{2} \\
\end{align}$
Taking minus sign we get the value of x as:
$\begin{align}
  & x=\dfrac{-5-7}{4} \\
 & \Rightarrow x=-\dfrac{12}{4}=-3 \\
\end{align}$
Hence, we have got the same values of x as that we have obtained in solution part of the question.