Question

Solve: $4{{x}^{2}}-3x-7$

Answer 1

Hint: In the question we are given a quadratic equation. To solve this we will use the method of factorisation or middle term splitting to simplify our given equation and then we will find the roots of our given equation.

Complete step by step answer:
In the question we are given a quadratic equation:
$4{{x}^{2}}-3x-7$
We will solve this quadratic equation using factorisation or middle term splitting
In the quadratic equation we will split middle term that is $-3x$ in two terms such that if we will add or subtract these terms we will get $-3x$ and if we multiply these terms then we will get product of other two terms $4{{x}^{2}}\times -7=-28{{x}^{2}}$
So we will split $-3x$ as $4x$ and $-7x$
If we add these two terms we will get $-3x$
$4x+(-7x)=-3x$
And if we multiply these two terms we will get $-28{{x}^{2}}$
$4x\times -7x=-28{{x}^{2}}$
Now we will replace $-3x$ with $4x-7x$
$-3x=4x-7x$
So our equation becomes.
$4{{x}^{2}}+4x-7x-7$
Now we will try to make factors of these equations.
$\begin{align}
  & 4{{x}^{2}}+4x-7x-7 \\
 & \Rightarrow 4x(x+1)-1(x+1) \\
 & \Rightarrow (4x-1)(x+1) \\
\end{align}$
If we substitute this equation equal to $0$
$4{{x}^{2}}-3x-7=0$
We can also write it as
$(4x-1)(x+1)=0$
Which will give us two conditions.
That are
Either $4x-1=0$ or $x+1=0$
If $4x-1=0$ then
$4x-1=0$
Transposing $1$ to right hand side
$\Rightarrow 4x=1$
Transposing $4$ to the right hand side.
$\Rightarrow x=\dfrac{1}{4}$
$\therefore x=\dfrac{1}{4}$ is the root of our given equation.
If $x+1=0$
Transposing $1$ to the right hand side.
$\Rightarrow x=-1$
$\therefore x=-1$ is the root of our given equation.
So there are two roots of our given quadratic equation and that are
$x=\dfrac{1}{4}$ And $x=-1$
Hence $x=\dfrac{1}{4}$ and $x=-1$ is our required solution.

Note:
A quadratic equation can have a maximum of two roots. There is a possibility that a quadratic equation has only one root or does not have any root. We can solve quadratic equations using three methods named as factorisation or middle term splitting, discriminant method and completing the square.