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# How do you solve the equation $2{{x}^{2}}+5x-3=0$ by using the quadratic formula?

Last updated date: 05th Aug 2024
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Hint: We have been given a quadratic equation of $x$ as $2{{x}^{2}}+5x-3=0$. We use the quadratic formula to solve the value of the $x$. we have the solution in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$. We put the values and find the solution.

Complete step-by-step solution:
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part $\sqrt{{{b}^{2}}-4ac}$ of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation.
In the given equation we have $2{{x}^{2}}+5x-3=0$. The values of a, b, c is $2,5,-3$ respectively.
We put the values and get $x$ as $x=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\times \left( -3 \right)\times 2}}{2\times 2}=\dfrac{-5\pm \sqrt{49}}{4}=\dfrac{-5\pm 7}{4}=-3,\dfrac{1}{2}$
The roots of the equation are real numbers.
So, values of x are $x=-3,\dfrac{1}{2}$.

Note: We can also apply the middle-term factoring or grouping to factorise the polynomial.
In the case of $2{{x}^{2}}+5x-3$, we break the middle term $5x$ into two parts of $6x$ and $-x$.
So, $2{{x}^{2}}+5x-3=2{{x}^{2}}+6x-x-3$. We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.
Here multiplication for both cases gives $-6{{x}^{2}}$. The grouping will be done for $2{{x}^{2}}+6x$ and $-x-3$.
We try to take the common numbers out.
For $2{{x}^{2}}+6x$, we take $2x$ and get $2x\left( x+3 \right)$.
For $-x-3$, we take $-1$ and get $-\left( x+3 \right)$.
The equation becomes $2{{x}^{2}}+5x-3=2{{x}^{2}}+6x-x-3=2x\left( x+3 \right)-\left( x+3 \right)$.
Both the terms have $\left( x+3 \right)$ in common. We take that term again and get
\begin{align} & 2{{x}^{2}}+5x-3 \\ & =2x\left( x+3 \right)-\left( x+3 \right) \\ & =\left( x+3 \right)\left( 2x-1 \right) \\ \end{align}
Therefore, $\left( x+3 \right)\left( 2x-1 \right)=0$ has multiplication of two polynomials giving a value of 0. This means at least one of them has to be 0.
Therefore, values of x are $x=-3,\dfrac{1}{2}$.