Question

How do you know how many solutions $2{{x}^{2}}+5x-7=0$ has?

Answer 1

Hint: From the question given we have to find the number of solutions of $2{{x}^{2}}+5x-7=0$
By using the quadratic discriminant formula we can find the number of solutions of the given quadratic equation $2{{x}^{2}}+5x-7=0$.If the discriminant is greater than $0$, that means that the quadratic equation has $2$ real solutions. If the discriminant is greater than $0$ and is a perfect square, the quadratic has two real and rational solutions. If the discriminant is exactly $0$, then the quadratic equation has exactly one real solution. If the discriminant is less than $0$, then the quadratic equation does not have any real solutions. The discriminant is given by the formulae ${{b}^{2}}-4ac$ .

Complete step by step answer:
Now let us evaluate the discriminant for our given quadratic equation,
Given quadratic equation is $2{{x}^{2}}+5x-7=0$
We can clearly observe that it is in the form of $a{{x}^{2}}+bx+c=0$
We also know that discriminant for the quadratic equation of the form $a{{x}^{2}}+bx+c=0$ is,
Discriminant is given by the formulae ${{b}^{2}}-4ac$.
By comparing the coefficients of the general form of the quadratic equation and the given equation from the question we get,
$\begin{align}
  & a=2 \\
 & b=5 \\
 & c=-7 \\
\end{align}$
Now, let us find the discriminant,
${{b}^{2}}-4ac={{\left( 5 \right)}^{2}}-4\left( 2 \right)\left( -7 \right)$
$\Rightarrow 25-\left( 8\left( -7 \right) \right)$
$\Rightarrow 25-\left( -56 \right)$
$\Rightarrow 81$
We can clearly observe that the discriminant is greater than zero, therefore the quadratic equation has two real solutions, and it is a perfect square, therefore the two solutions are also rational.

Therefore, the given equation has $2$ real and rational solutions.

Note: We should be well aware of the conditions of the discriminant. We should be well aware of the applying discriminant formula for the quadratic equation to know how many solutions the given equation has. We can also this question by directly finding the zeros of this equation $2{{x}^{2}}+5x-7=0$ using the formulae $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ we will have $\dfrac{-5\pm \sqrt{25-4\left( 2 \right)\left( -7 \right)}}{2\left( 2 \right)}=\dfrac{-5\pm \sqrt{25+56}}{4}=\dfrac{-5\pm \sqrt{81}}{4}=\dfrac{-5\pm 9}{4}=\dfrac{-7}{2},1$ so there are 2 distinct real solutions.