Answer

Verified

388.2k+ views

**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 $x=-3,\dfrac{1}{2}$.

Recently Updated Pages

Which of the following is correct regarding the Indian class 10 social science CBSE

Who was the first sultan of delhi to issue regular class 10 social science CBSE

The Nagarjuna Sagar project was constructed on the class 10 social science CBSE

Which one of the following countries is the largest class 10 social science CBSE

What is Biosphere class 10 social science CBSE

Read the following statement and choose the best possible class 10 social science CBSE

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Difference Between Plant Cell and Animal Cell

10 examples of law on inertia in our daily life