Key Takeaways
Over the real numbers, a square cannot equal a negative number. Completing the square exposes why a quadratic has no real root, one distinct repeated root or two distinct real roots, so the discriminant need not remain a memorized rule.
- State a≠0 and the real-number scope before saying there is no real root.Read the supporting section: State the equation type and number system
- Squares equal to -1, 0 and 1 give zero, one and two distinct real roots.Read the supporting section: The same square with three right-hand sides
- Substitute each candidate; the no-root proof uses nonnegative squares, not a finite search.Read the supporting section: Substitute back and identify the missing condition
In this article
State the equation type and number system
OpenStax’s quadratic-function discussion connects parabolas, zeros and horizontal-axis intersections. Here ax²+bx+c=0 has a nonzero a and real-valued solutions. If a=0, return to the resulting linear or other case instead of continuing to apply a quadratic classification.
No real root does not mean no solution in every number system. Complex solutions are outside this lesson. Writing the number system first prevents answers from different course contexts from appearing contradictory.
The same square with three right-hand sides
Since x²+4x=(x+2)²-4, the three constants 5, 4 and 3 produce squares equal to -1, 0 and 1. Preserve equality at every step; adding 4 on only one side is not a valid completion of the square.
| Equation | Completed square | Real solution set |
|---|---|---|
| x²+4x+5=0 | (x+2)²=-1 | Empty |
| x²+4x+4=0 | (x+2)²=0 | {-2} |
| x²+4x+3=0 | (x+2)²=1 | {-3, -1} |
Explain the count instead of copying a set
In the first case, no real square equals -1. In the second, x+2 must be 0, so x=-2. In the third, consider both x+2=1 and x+2=-1, giving -1 and -3. Omitting the negative branch loses a valid solution.
The discriminants are -4, 0 and 4, respectively. Relate that familiar classification to the square argument you just established. Algebraic multiplicity and the number of distinct solutions are different counts; ±0 does not create two distinct values of -2.
Substitute back and identify the missing condition
In the third equation, -3 gives 9-12+3=0 and -1 gives 1-4+3=0. In the second, -2 gives 4-8+4=0. Trying a few values does not prove the first has no real root; the proof is that (x+2)²+1 is always at least 1.
For a different reason candidates can be excluded, see the rational-equation domain hole and original logarithmic domain. This lesson concerns the range of a real square rather than invalid roots introduced by cancellation.
For independent practice, replace 4x with 6x and choose constants 10, 9 and 8. Obtain (x+3)²=-1, 0 and 1, with solution sets empty, {-3} and {-4, -2}. Explain all three counts and substitute each candidate to demonstrate the method.
If a change makes a = 0, recognize that the equation is no longer quadratic; if the course changes to complex numbers, a separate discussion is needed. Check the number system and highest-degree coefficient before computing a discriminant or counting roots.
