Question

If $2{{x}^{2}}-5x-3=0$ , then find $x$ .

Answer 1

Hint: We need to find the roots of the equation $2{{x}^{2}}-5x-3=0$ . We will be using the formula
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . For this, first we will find discriminant, $D={{b}^{2}}-4ac$ and substitute that value in the equation $x=\dfrac{-b\pm \sqrt{D}}{2a}$ . By suitable substitutions and splitting the $\pm $ sign , we will get the values of $x$ .

Complete step by step answer:
We need to find the roots of the equation $2{{x}^{2}}-5x-3=0$ .
Let us use the discriminant in factorizing this.
We know that $D={{b}^{2}}-4ac$ for an equation $a{{x}^{2}}+bx+c=0$ .
Now, let us compare this with $2{{x}^{2}}-5x-3=0$ .
By doing so, we get $a=3,b=-5,c=-3$
Thus, $D={{(-5)}^{2}}-4\times 2\times (-3)$
Simplifying this gives
$D=25+24=49$
We have studied that, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ or
$x=\dfrac{-b\pm \sqrt{D}}{2a}$
Let us now substitute the values.
$x=\dfrac{-(-5)\pm \sqrt{49}}{2\times 2}$
Simplifying this gives
$x=\dfrac{5\pm 7}{4}$
This can be written as
$x=\dfrac{5+7}{4},\dfrac{5-7}{4}$
Solving, we get
$x=\dfrac{12}{4},\dfrac{-2}{4}$
Simplifying this further, we get
$x=3,-\dfrac{1}{2}$
Hence the value of $x=3,-\dfrac{1}{2}$ .

Note:
The discriminant need not be found in the beginning. You can directly substitute in the equation $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Be careful with the $\pm $ sign and do not miss to write it as it is the one that gives two values to $x$. Since the given polynomial is of degree $2$, there will be two values for $x$. This question can also be solved as follows:
Given, $2{{x}^{2}}-5x-3=0$
Diving throughout by $2$, we get
${{x}^{2}}-\dfrac{5}{2}x-\dfrac{3}{2}=0$
We can use factorization by splitting the middle term.
Now, the denominator of the second and third term is same. Hence,
$-3\times \dfrac{1}{2}=\dfrac{-3}{2}$ and $-3+\dfrac{1}{2}=-\dfrac{5}{2}$ .
Thus ${{x}^{2}}+\dfrac{1}{2}x-3x-\dfrac{3}{2}=0$
Taking, the common terms outside, we get
$x\left( x+\dfrac{1}{2} \right)-3\left( x+\dfrac{1}{2} \right)=0$
This, can be written as
$\left( x-3 \right)\left( x+\dfrac{1}{2} \right)=0$
Equating each term to $0$ we get
$\Rightarrow \left( x-3 \right)=0,\left( x+\dfrac{1}{2} \right)=0$
Consider $\left( x-3 \right)=0$
By solving this, we get
$\Rightarrow x=3$
Consider $\left( x+\dfrac{1}{2} \right)=0$
By solving this, we get
$\Rightarrow x=-\dfrac{1}{2}$
Hence the value of $x=3,-\dfrac{1}{2}$ .