HOW MANY DISTINCT REAL ROOTS DOES x4 + 3x2 − 4 = 0 HAVE?
There are two distinct real roots, x = 1 and x = −1. Substituting u = x² gives u = 1 and u = −4, and only the non-negative value of u can come from a real x, so the pair of roots from u = −4 does not exist.
How many distinct real roots does the equation x4 + 3x2 − 4 = 0 have?
It is a quadratic in x². Solve it as one, then filter the solutions before you count them, because a negative value of x² produces no real x at all.
- put u = x2: u2 + 3u − 4 = 0
- (u + 4)(u − 1) = 0, so u = −4 or u = 1
- u = −4 is rejected: x2 cannot be negative for real x
- u = 1 gives x = ±1
- that is 2 distinct real roots
4 is what you get if you take square roots of both values of u and treat √(−4) as though it produced real roots. A quartic has four roots counted with multiplicity over the complex numbers, and the question asks only for the real ones. The option is there for anyone who counts before filtering.
1 is what you get if you solve u = 1 and stop at x = 1 forgetting that x² = 1 has two solutions. Every positive value of u contributes a pair.
You can count without solving. Each positive root in u gives two real roots in x, u = 0 gives one, and each negative root gives none. So sketch or sign-check the quadratic in u, count its positive roots, double that, and you have the answer. Here the product of the u roots is −4, which is negative, so one root is positive and one is negative: one positive root, two real x, done.
THE GENERAL RULE
Any equation in x⁴ and x² only — and the same goes for x⁶ and x³, or sin²θ and sinθ — reduces to a quadratic. The step the question is actually testing is the one after the quadratic: converting back, and checking whether each solution is admissible in the original variable. Counting the roots of the substituted equation and reporting that number is the standard error.
Updated · written to the published ESAT content specification