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Unit 6: Quadratic and Polynomial Functions

Topic 3

Quadratic Formula and Discriminant

When a quadratic equation cannot be easily factored, we use the quadratic formula:
x=−b±b2−4ac2a.x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
It gives the exact solutions to any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 where a≠0a \ne 0.
The expression under the square root,
Δ=b2−4ac,\Delta = b^2 - 4ac,
is called the discriminant. It determines the nature and number of the roots.
Meaning of the Discriminant
  • Δ>0:\Delta > 0: two distinct real roots
  • Δ=0:\Delta = 0: one real root (a repeated or double root)
  • Δ<0:\Delta < 0: no real roots (two complex solutions)
On the SAT, problems often test whether you can use the discriminant to reason about the number or type of solutions without computing them.

Core Skills

  • Identify a,b,ca, b, c in standard form.
  • Substitute into the quadratic formula correctly.
  • Simplify radicals accurately.
  • Interpret the discriminant to determine the number of real solutions.
  • Apply the formula to both exact and approximate contexts.

Example 1: Using the Formula Directly

Solve x2−5x+6=0.x^2 - 5x + 6 = 0. Here a=1,  b=−5,  c=6.a = 1,\; b = -5,\; c = 6.
x=−(−5)±(−5)2−4(1)(6)2(1)=5±25−242=5±12.x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}.
x=3 or 2.x = 3 \text{ or } 2.
Solutions: x=2,3\boxed{x = 2, 3}

Example 2: No Real Roots

Solve x2+4x+8=0.x^2 + 4x + 8 = 0. a=1,  b=4,  c=8.a = 1,\; b = 4,\; c = 8.
x=−4±42−4(1)(8)2(1)=−4±16−322=−4±−162.x = \frac{-4 \pm \sqrt{4^2 - 4(1)(8)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 32}}{2} = \frac{-4 \pm \sqrt{-16}}{2}.
Since the discriminant is negative, no real roots exist. Answer: No real solutions.\boxed{\text{No real solutions.}}

Example 3: One Double Root

Solve x2−6x+9=0.x^2 - 6x + 9 = 0. a=1,  b=−6,  c=9.a = 1,\; b = -6,\; c = 9.
Δ=(−6)2−4(1)(9)=36−36=0.\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0.
x=−(−6)2(1)=62=3.x = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3.
One double root: x=3\boxed{x = 3}

Example 4: Simplifying with Square Roots

Solve 2x2−3x−2=0.2x^2 - 3x - 2 = 0. a=2,  b=−3,  c=−2.a = 2,\; b = -3,\; c = -2.
x=−(−3)±(−3)2−4(2)(−2)2(2)=3±9+164=3±254=3±54.x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4}.
x=2 or −12.x = 2 \text{ or } -\tfrac{1}{2}.
Solutions: x=2,−12\boxed{x = 2, -\tfrac{1}{2}}

Example 5: Interpreting the Discriminant

For 4x2+4x+1=0,4x^2 + 4x + 1 = 0,
Δ=42−4(4)(1)=16−16=0.\Delta = 4^2 - 4(4)(1) = 16 - 16 = 0.
So there is one repeated real solution. Root: x=−48=−12.x = \frac{-4}{8} = -\tfrac{1}{2}.

Example 6: SAT Application

A projectile’s height is given by h=−16t2+32t+48.h = -16t^2 + 32t + 48. When does it hit the ground (h=0h = 0)?
−16t2+32t+48=0⇒t=−32±322−4(−16)(48)2(−16).-16t^2 + 32t + 48 = 0 \Rightarrow t = \frac{-32 \pm \sqrt{32^2 - 4(-16)(48)}}{2(-16)}.
t=−32±1024+3072−32=−32±4096−32=−32±64−32.t = \frac{-32 \pm \sqrt{1024 + 3072}}{-32} = \frac{-32 \pm \sqrt{4096}}{-32} = \frac{-32 \pm 64}{-32}.
t=3 or −1.t = 3 \text{ or } -1.
Time cannot be negative, so t=3 seconds\boxed{t = 3\text{ seconds}}.

Key Takeaways

  • The quadratic formula works for all quadratics—factorable or not.
  • The discriminant b2−4acb^2 - 4ac determines the number and type of roots.
  • Always simplify radicals carefully and reduce fractions.
  • If Δ=0\Delta = 0, expect one double root.
  • If Δ<0\Delta < 0, the solutions are complex (no real roots on SAT).