By Tricia Lobo, Updated Aug 30, 2022
In algebra, the phrase “all real solutions” means you should determine every value that satisfies the equation, ignoring any complex results that involve the imaginary unit i. The strategy is identical for equations that yield only real numbers and those that produce both real and complex solutions: solve the equation, then discard any non‑real answers.
Reduce the expression to its simplest form. For example, if you have x^4 + x^2 – 6 = 0, use the substitution u = x^2 to obtain u^2 + u – 6 = 0. This makes the equation easier to factor.
Rewrite the quadratic in terms of u and factor it. Continuing the example, we can express the left‑hand side as
u^2 + 3u – 2u – 6 = 0\n\t= u(u + 3) – 2(u + 3) = (u – 2)(u + 3) = 0.
Set each factor equal to zero. Here, u – 2 = 0 gives u = 2 and u + 3 = 0 gives u = –3. Since u = x^2, the corresponding real solutions are x = ±√2 and x = ±√3 (the negative root of u = –3 yields an imaginary number, so it is discarded).
Any root that involves the square root of a negative number is complex and should be excluded from the final list of real solutions. In this example, all solutions are real, so no discard is necessary.
