SAT Math – Advanced Math
Solving Quadratic Equations
Finding solutions using factoring, quadratic formula, and completing the square
Solving quadratic equations means finding the values of x that make \(ax^2 + bx + c = 0\) true. On the SAT, you'll solve quadratics by factoring, using the quadratic formula, completing the square, or graphing—each method suited to different situations and providing insights into parabola behavior.
Success requires recognizing which method is most efficient, understanding the Zero Product Property, interpreting the discriminant to determine the number of solutions, and connecting algebraic solutions to x-intercepts on graphs. These techniques aren't just equation-solving mechanics—they're fundamental to analyzing projectile motion, optimizing profit functions, finding break-even points, and modeling any phenomenon involving squared variables.
Understanding Quadratic Equations
Standard Form and Solutions
Standard form: \(ax^2 + bx + c = 0\) where \(a \neq 0\)
Also called: Zeros, x-intercepts, roots
Number of solutions: 0, 1, or 2 real solutions
Graphically: Where parabola crosses x-axis
Solving by Factoring
Factor the quadratic and use the Zero Product Property.
Process: Factor, set each factor = 0, solve
Example: \(x^2 - 5x + 6 = 0\) → \((x-2)(x-3) = 0\) → \(x = 2\) or \(x = 3\)
Best when: Quadratic factors easily
The Quadratic Formula
Universal method that always works for any quadratic equation.
Discriminant: \(b^2 - 4ac\) determines number of solutions
Use when: Factoring is difficult or impossible
Always works: Guaranteed to find all solutions
Discriminant Analysis
The expression \(b^2 - 4ac\) reveals information about solutions.
If \(b^2 - 4ac = 0\): One real solution (repeated root)
If \(b^2 - 4ac < 0\): No real solutions (complex solutions)
Graphically: Crosses x-axis twice, touches once, or doesn't cross
Essential Formulas and Methods
The Quadratic Formula
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
For equation \(ax^2 + bx + c = 0\)
± means two solutions (use + for one, − for the other)
Memorize this formula—it's not provided on the SAT!
Factoring Method
For \(x^2 + bx + c = 0\):
1. Find two numbers that multiply to c and add to b
2. Write as \((x + m)(x + n) = 0\)
3. Set each factor equal to zero and solve
Zero Product Property: If \(AB = 0\), then \(A = 0\) or \(B = 0\)
Completing the Square
For \(x^2 + bx + c = 0\):
1. Move c to right side: \(x^2 + bx = -c\)
2. Add \(\left(\frac{b}{2}\right)^2\) to both sides
3. Factor left side as perfect square
4. Take square root and solve
Special Cases
Perfect squares: \(x^2 = k\) → \(x = \pm\sqrt{k}\)
Difference of squares: \(x^2 - k = 0\) → \((x-\sqrt{k})(x+\sqrt{k}) = 0\)
Missing b term: \(ax^2 + c = 0\) → isolate \(x^2\) and take square root
Common Pitfalls & Expert Tips
❌ Forgetting to set equation equal to zero first
Can't factor or use quadratic formula unless equation is in form \(ax^2 + bx + c = 0\). Move all terms to one side first!
❌ Missing the ± when taking square roots
If \(x^2 = 16\), the answer is \(x = \pm 4\) (both +4 and -4). Don't forget the negative solution!
❌ Sign errors in quadratic formula
Formula is \(-b\), not just b! If \(b = -5\), then \(-b = -(-5) = +5\). Also, \(-4ac\) has a negative already!
❌ Dividing by x or canceling incorrectly
From \(x^2 = 4x\), don't divide both sides by x! You'll lose the \(x = 0\) solution. Factor: \(x^2 - 4x = 0\) → \(x(x-4) = 0\).
✓ Expert Tip: Try factoring first
Factoring is fastest when it works. Try it first—if factors don't appear quickly, switch to quadratic formula.
✓ Expert Tip: Check your answers
Substitute solutions back into original equation. Both should make it true. Quick verification catches arithmetic errors!
✓ Expert Tip: Use discriminant strategically
Before solving, check \(b^2 - 4ac\). If negative, no real solutions exist—saves time! If perfect square, equation factors nicely.
Fully Worked SAT-Style Examples
Solve: \(x^2 + 7x + 12 = 0\)
Solution:
Step 1: Find factors of 12 that add to 7
Factor pairs of 12: 1×12, 2×6, 3×4
3 + 4 = 7 ✓ and 3 × 4 = 12 ✓
Step 2: Write factored form
\((x + 3)(x + 4) = 0\)
Step 3: Apply Zero Product Property
\(x + 3 = 0\) or \(x + 4 = 0\)
\(x = -3\) or \(x = -4\)
Check:
For \(x = -3\): \((-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0\) ✓
For \(x = -4\): \((-4)^2 + 7(-4) + 12 = 16 - 28 + 12 = 0\) ✓
Answer: \(x = -3\) or \(x = -4\)
Solve: \(2x^2 + 5x - 3 = 0\)
Solution:
Identify coefficients:
\(a = 2\), \(b = 5\), \(c = -3\)
Apply quadratic formula:
\(x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-3)}}{2(2)}\)
\(= \frac{-5 \pm \sqrt{25 + 24}}{4}\)
\(= \frac{-5 \pm \sqrt{49}}{4}\)
\(= \frac{-5 \pm 7}{4}\)
Find both solutions:
\(x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}\)
\(x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\)
Answer: \(x = \frac{1}{2}\) or \(x = -3\)
Solve: \(x^2 - 25 = 0\)
Solution:
Method 1: Isolate \(x^2\)
\(x^2 = 25\)
\(x = \pm\sqrt{25} = \pm 5\)
Method 2: Factor as difference of squares
\(x^2 - 25 = (x + 5)(x - 5) = 0\)
\(x + 5 = 0\) or \(x - 5 = 0\)
\(x = -5\) or \(x = 5\)
Answer: \(x = \pm 5\) (or \(x = 5\) or \(x = -5\))
Solve: \(x^2 = 3x + 10\)
Solution:
Step 1: Move all terms to one side
\(x^2 - 3x - 10 = 0\)
Step 2: Factor
Need factors of -10 that add to -3: -5 and 2
\((x - 5)(x + 2) = 0\)
Step 3: Solve
\(x = 5\) or \(x = -2\)
Common Error:
Don't divide both sides by x!
From \(x^2 = 3x + 10\), dividing by x loses the solution \(x = -2\)
Answer: \(x = 5\) or \(x = -2\)
How many real solutions does \(x^2 + 4x + 5 = 0\) have?
Solution:
Use discriminant: \(b^2 - 4ac\)
\(a = 1\), \(b = 4\), \(c = 5\)
\(b^2 - 4ac = 4^2 - 4(1)(5)\)
\(= 16 - 20 = -4\)
Interpret result:
Discriminant = -4 (negative)
Negative discriminant → No real solutions
Graphical interpretation:
Parabola opens upward (\(a = 1 > 0\))
Never crosses x-axis (no x-intercepts)
Answer: Zero real solutions (no real solutions)
Solve: \(x^2 - 6x + 9 = 0\)
Solution:
Recognize perfect square trinomial:
First term: \(x^2 = (x)^2\)
Last term: \(9 = (3)^2\)
Middle: \(2 \times x \times 3 = 6x\) ✓
Factor as perfect square:
\((x - 3)^2 = 0\)
or \((x - 3)(x - 3) = 0\)
Solve:
\(x - 3 = 0\)
\(x = 3\) (repeated root)
Note:
Discriminant: \((-6)^2 - 4(1)(9) = 36 - 36 = 0\)
Zero discriminant → one solution (vertex touches x-axis)
Answer: \(x = 3\)
Solve: \(3x^2 - 12x = 0\)
Solution:
Step 1: Factor out GCF
GCF of \(3x^2\) and \(12x\) is 3x
\(3x(x - 4) = 0\)
Step 2: Apply Zero Product Property
\(3x = 0\) or \(x - 4 = 0\)
\(x = 0\) or \(x = 4\)
Don't miss x = 0!
Zero is a valid solution
Always factor out GCF first to avoid losing solutions
Answer: \(x = 0\) or \(x = 4\)
The height h (in feet) of a ball t seconds after being thrown is modeled by \(h = -16t^2 + 32t + 48\). When does the ball hit the ground?
Solution:
Set height equal to zero:
Ground level means \(h = 0\)
\(-16t^2 + 32t + 48 = 0\)
Simplify by dividing by -16:
\(t^2 - 2t - 3 = 0\)
Factor:
Factors of -3 that add to -2: -3 and 1
\((t - 3)(t + 1) = 0\)
\(t = 3\) or \(t = -1\)
Context interpretation:
Time cannot be negative
Discard \(t = -1\)
Ball hits ground at t = 3 seconds
Answer: 3 seconds
Method Selection Guide
Use Factoring When:
• Factors are obvious
• Leading coefficient is 1
• Perfect square or difference of squares
Use Quadratic Formula When:
• Factoring is difficult
• Coefficients are large
• Need exact decimal answers
Solving Quadratic Equations: Gateway to Advanced Mathematics
Quadratic equations represent the next level of mathematical sophistication beyond linear relationships—they capture acceleration, curvature, and optimization in ways straight lines cannot. The SAT tests these solving skills because they're indispensable across quantitative fields: physicists solve quadratics to find projectile landing times and maximum heights, engineers use them to optimize product dimensions and minimize costs, economists apply them to find profit-maximizing production levels, and computer scientists employ them in algorithm complexity analysis. Master multiple solution methods because efficiency matters—factoring provides instant solutions when patterns are clear, the quadratic formula guarantees results when factoring fails, and understanding the discriminant reveals solution behavior before calculation. The Zero Product Property isn't just a solving technique but a profound insight: products equal zero only when factors equal zero, connecting multiplication to roots in fundamental ways. These solutions aren't abstract—they're x-intercepts where parabolas cross the axis, break-even points where profit equals zero, times when objects reach ground level, and values where functions achieve specific targets. The fluency you develop—recognizing perfect squares instantly, applying the quadratic formula flawlessly, checking discriminants strategically—distinguishes students who merely follow procedures from those who understand quadratic structure deeply. Every time you solve \(ax^2 + bx + c = 0\), you're finding where a parabola intersects a horizontal line, revealing equilibrium points in dynamic systems governed by squared relationships.
