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# What is the nature of roots of the quadratic equation $2{{x}^{2}}-\sqrt{5}x+1=0$. Verified
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Hint: In this problem, we have to find the nature of the roots of the quadratic equation. We should know that the general quadratic equation is of the form $a{{x}^{2}}+bx+c=0$, where a, b, c are constant terms and a is not equal to zero. we can compare the given equation and the general equation for the value of a, b, c to be substituted in the formula $\Delta ={{b}^{2}}-4ac$ and find the nature of the roots.

Here we can see the formula to solve the general form of the quadratic equation and what its discriminant value is.
We know that the general quadratic equation is of the form,
$a{{x}^{2}}+bx+c=0$……. (1)
We know that the given equation is,
$2{{x}^{2}}-\sqrt{5}x+1=0$……. (2)
By comparing (1) and (2), we get a = 2, b = $-\sqrt{5}$, c = 1.
Where, the discriminant is denoted as $\Delta$,
$\Delta ={{b}^{2}}-4ac$ …… (3)
We should know that,
If the discriminant value is equal 0, $\Delta =0$
Then we will have two real and equal roots (i.e. one real root).
If the discriminant value is greater than 0, $\Delta >0$
Then we will have two real and unequal roots.
If the discriminant value is less than 0, $\Delta <0$
Then we will have no real roots (imaginary roots).
We can now substitute the values in (3), we get
\begin{align} & \Rightarrow \Delta ={{\left( -\sqrt{5} \right)}^{2}}-4\left( 2 \right)=5-8 \\ & \Rightarrow \Delta =-3<0 \\ \end{align}
Therefore, there will be no real roots.

Note: We should always remember the condition of the discriminants which tells the type of roots, when the discriminant is equal to zero, the equation has one real root, when the discriminant is less than 0, it has no real roots and when the discriminant is greater than 0, then it has two real roots.