The Discriminant & the Sum and Product of Roots

The discriminant: counting roots before you find them

Look back at the quadratic formula. The action all happens under the square root, in the expression b^2 - 4ac. That quantity is the [[discriminant|discriminant]], written D. Because you cannot take the real square root of a negative number, the *sign* of D alone decides how many real roots the equation has — no full solving required.

1. If D > 0: two distinct real roots — the parabola crosses the x-axis twice.
2. If D = 0: one repeated real root, a double root — the parabola just touches the x-axis.
3. If D < 0: no real roots — the parabola never meets the x-axis (the roots are complex).

Discriminant  D = b^2 - 4ac.

  x^2 - 5x + 6 = 0:  D = 25 - 24 = 1 > 0  → 2 real roots
  x^2 - 6x + 9 = 0:  D = 36 - 36 = 0      → 1 double root (x = 3)
  x^2 + x + 1 = 0:   D = 1 - 4  = -3 < 0  → no real roots

A perfect-square D (like 1, 4, 9, ...) is a bonus signal:
the roots are rational, so the quadratic factors nicely.

Sum and product of roots: Vieta's formulas

There is a beautiful shortcut linking the two roots back to the coefficients without solving at all. If r and s are the roots of a x^2 + b x + c = 0, then their [[sum-and-product-of-roots|sum and product]] obey [[vietas-formulas|Vieta's formulas]]: r + s = -b/a and r·s = c/a. You can see why instantly — expanding a(x - r)(x - s) and matching coefficients with a x^2 + b x + c forces exactly these relations.

Vieta's formulas for a x^2 + b x + c = 0 with roots r, s:

    r + s = -b/a       r·s = c/a

Check on  2x^2 + 3x - 5 = 0  (roots were 1 and -5/2):

    r + s = 1 + (-5/2) = -3/2 = -b/a = -3/2   ✓
    r·s   = 1·(-5/2)   = -5/2 = c/a  = -5/2   ✓

Use it backward — build a quadratic with roots 3 and -4:
    sum = -1,  product = -12
    x^2 - (sum)x + (product) = x^2 + x - 12 = 0