Find the number of real roots of the equation 2x^4 + 5x^2 + 3 = 0

We have the equation 2x^4 + 5x^2 + 3 = 0

To convert this into a quadratic equation, we'll assume x^2  = t

The equation will now be 2t^2 + 5t + 3 = 0

The question can now be solved using two approaches. One is complete solving and the other is intuitive.

COMPLETE SOLVING - We have 

2t^2 + 5t + 3 = 0

=> 2t^2 + 2t + 3t + 3 = 0

=> 2t(t + 1) + 3(t + 1) = 0

=> (t + 1)(2t + 3) = 0

=> t = -1, -3/2

We've had assumed x^2 = t. So, 

x^2  = -1; x^2 = -3/2

From these results, we won't get any real real value of x. Or in other words, any real value of x will not satisfy these results because Square of any real number is always greater than or equal to 0. So,2x^4 + 5x^2 + 3 = 0 will have no real roots. Or we can say that the number of real roots of 2x^4 + 5x^2 + 3 = 0 is 0

INTUITIVE APPROACH - We have

2t^2 + 5t + 3 = 0

Sum of roots = -5/2
Product of roots = 3/2

Here, Sum of roots is negative (<0) and product of roots is positive (>0). This gives us clear indication that both the roots are negative (<0).

Since both values of t are negative, both values of x^2 are negative, which is obviously not possible for any real value of x. So the equation 2x^4 + 5x^2 + 3 = 0 will have no real roots.