Level 6 · Ages 16–17
Read it again any time, including while you work through the exercises.
Throw a ball into the air: its path draws an arched curve, a parabola. This curve is the graph of a quadratic function, of the form f(x) = ax² + bx + c, with a nonzero. The x² term, new compared with linear functions, is responsible for the bending.
Solving ax² + bx + c = 0 means finding where the parabola crosses the horizontal axis. Depending on its position, it can cut it twice, touch it exactly once, or stay entirely above or below. So there are zero, one or two solutions.
A magic number settles the question: the discriminant, written Δ (delta), with Δ = b² - 4ac. If Δ > 0, there are two solutions: x = (-b - √Δ) ÷ (2a) and x = (-b + √Δ) ÷ (2a). If Δ = 0, exactly one solution: x = -b ÷ (2a). If Δ < 0, no real solution: you cannot take the square root of a negative number.
Traps to avoid: identify a, b and c with their signs (in x² - 3x + 2, we have b = -3), do not forget the 2a in the denominator (2a, not 2), and remember that -b flips the sign of b. Finally, the sign of a gives the parabola its shape: branches up if a > 0, down if a < 0.
Solve x² - 5x + 6 = 0.
a = 1, b = -5, c = 6
I identify the coefficients with their signs: careful, b equals -5.
Δ = (-5)² - 4 × 1 × 6 = 25 - 24 = 1
I compute the discriminant. Δ = 1 > 0: there will be two solutions.
x = (5 - 1) ÷ 2 = 2 et x = (5 + 1) ÷ 2 = 3
I apply the formula x = (-b ± √Δ) ÷ (2a): here -b = 5 and √Δ = 1.
2² - 5 × 2 + 6 = 0 et 3² - 5 × 3 + 6 = 0
Check: 4 - 10 + 6 = 0 and 9 - 15 + 6 = 0. Both solutions are confirmed.
Result: The solutions are x = 2 and x = 3.
Solve x² + 2x + 5 = 0.
a = 1, b = 2, c = 5
I identify the coefficients.
Δ = 2² - 4 × 1 × 5 = 4 - 20 = -16
The discriminant is negative.
Δ < 0
No real number has a negative square, so √Δ does not exist: the equation has no real solution. The parabola never touches the horizontal axis.
Result: No real solution.