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Quadratic Solutions

Find sums, products, and the number of solutions without solving the equation.

Mastery
Advanced Math

Product of Solutions

\[ \text{Product} = \frac{c}{a} \]

INSTANTLY find what the solutions multiply to. Essential when the SAT asks for the product of roots.

Practice: "Product of solutions to \(3x^2 - 7x + 12 = 0\)?"

Show Solution & Analysis
🚫 School Way (Rigorous)

1. Setup Quadratic Formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Identify \(a=3, b=-7, c=12\).
\(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(12)}}{2(3)}\)
2. Simplify:
\(x = \frac{7 \pm \sqrt{49 - 144}}{6} = \frac{7 \pm \sqrt{-95}}{6}\)
3. Define Complex Roots:
\(x_1 = \frac{7 + i\sqrt{95}}{6}, x_2 = \frac{7 - i\sqrt{95}}{6}\)
4. Calculate Product:
\(x_1 \cdot x_2 = \left(\frac{7 + i\sqrt{95}}{6}\right)\left(\frac{7 - i\sqrt{95}}{6}\right)\)
Numerator: \((7+i\sqrt{95})(7-i\sqrt{95}) = 49 - (i\sqrt{95})^2 = 49 - (-95) = 144\).
Denominator: \(6 \cdot 6 = 36\).
5. Result: \(144 / 36 = 4\).

✅ Practix Way (Optimal)

**Step 1:** Identify coefficients.
\(a=3, c=12\).
**Step 2:** Apply shortcut.
Product = \(c/a = 12/3\).
**Result: 4**
Time saved: 85+ seconds.

🎓 Theoretical Solution: The Derivation

Any quadratic equation can be written in Standard Form:

\( ax^2 + bx + c = 0 \)

Dividing by \(a\) gives the Monical Form:

\( x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \)

If \(x_1\) and \(x_2\) are the roots, the equation also factors as:

\( (x - x_1)(x - x_2) = x^2 - (x_1 + x_2)x + x_1x_2 = 0 \)

By matching the constant terms of both expressions, we find:

\( x_1x_2 = \frac{c}{a} \)
Mastery
Advanced Math

Sum of Solutions

\[ \text{Sum} = -\frac{b}{a} \]

Instantly find what the solutions add up to. Don't waste time solving for individual roots.

Practice: "Sum of solutions to \(2x^2 + 8x - 5 = 0\)?"

Show Solution & Analysis
🚫 School Way (Rigorous)

1. Setup Quadratic Formula:
Identify \(a=2, b=8, c=-5\).
\(x = \frac{-8 \pm \sqrt{8^2 - 4(2)(-5)}}{2(2)}\)
2. Simplify Discriminant:
\(x = \frac{-8 \pm \sqrt{64 + 40}}{4} = \frac{-8 \pm \sqrt{104}}{4}\)
3. Simplify Terms:
\(x = \frac{-8}{4} \pm \frac{\sqrt{104}}{4} = -2 \pm \frac{2\sqrt{26}}{4} = -2 \pm \frac{\sqrt{26}}{2}\)
4. Define Roots:
\(x_1 = -2 + \frac{\sqrt{26}}{2}, x_2 = -2 - \frac{\sqrt{26}}{2}\)
5. Calculate Sum:
\(x_1 + x_2 = (-2 + \frac{\sqrt{26}}{2}) + (-2 - \frac{\sqrt{26}}{2})\)
Radicals cancel out: \(-2 - 2 = -4\).
Result: -4

✅ Practix Way (Optimal)

**Step 1:** Identify coefficients.
\(a=2, b=8\).
**Step 2:** Apply shortcut.
Sum = \(-b/a = -8/2\).
**Result: -4**
Done in seconds.

🎓 Theoretical Solution: The Derivation

Starting from Standard Form:

\( ax^2 + bx + c = 0 \)

Dividing by \(a\) reveals individual coefficient contributions:

\( x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \)

Using the Factor Theorem with roots \(x_1\) and \(x_2\):

\( (x - x_1)(x - x_2) = x^2 - (x_1 + x_2)x + x_1x_2 = 0 \)

By matching the linear coefficient (\(x\)-term), we see:

\( -(x_1 + x_2) = \frac{b}{a} \implies x_1 + x_2 = -\frac{b}{a} \)
Mastery
Advanced Math

Difference of Solutions

\[ \text{Diff} = \pm \frac{\sqrt{b^2 - 4ac}}{a} \]

Find the difference between the two solutions without having to find each root individually first.

Practice: "Difference of solutions to \(x^2 - 4x - 5 = 0\)?"

Show Solution & Analysis
🚫 School Way (Rigorous)

1. Setup Quadratic Formula:
Identify \(a=1, b=-4, c=-5\).
\(x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}\)
2. Simplify Discriminant:
\(x = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2}\)
3. Find Both Roots:
\(x_1 = \frac{4 + 6}{2} = 5, \quad x_2 = \frac{4 - 6}{2} = -1\)
4. Subtract the Roots:
\( |5 - (-1)| = |5 + 1| = 6 \)
Result: 6

✅ Practix Way (Optimal)

Step 1: Compute Discriminant \(D\).
\(D = (-4)^2 - 4(1)(-5) = 36\).
Step 2: Apply shortcut.
Difference = \( \frac{\sqrt{D}}{|a|} = \frac{\sqrt{36}}{|1|} = 6\).
Result: 6
Done in seconds.

🎓 Theoretical Solution: The Derivation

We want to find \(|x_1 - x_2|\). Notice that:

\( (x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2 \)

We already know the shortcuts for Sum and Product:

  • Sum: \( x_1 + x_2 = -\frac{b}{a} \)
  • Product: \( x_1x_2 = \frac{c}{a} \)

Substitute these into our equation:

\( (x_1 - x_2)^2 = \left(-\frac{b}{a}\right)^2 - 4\left(\frac{c}{a}\right) \)
\( (x_1 - x_2)^2 = \frac{b^2}{a^2} - \frac{4ac}{a^2} = \frac{b^2 - 4ac}{a^2} \)

Finally, take the square root of both sides to get the absolute difference:

\( |x_1 - x_2| = \frac{\sqrt{b^2 - 4ac}}{|a|} \)
Mastery
Advanced Math

The Discriminant

\[ D = b^2 - 4ac \]

\(D > 0\): 2 Real Solutions
\(D = 0\): 1 Real Solution
\(D < 0\): No Real Solutions

Practice: "How many real solutions for the equation \(x^2 + 4x + 4 = 0\)?"

Show Solution & Analysis
🚫 School Way (Rigorous)

1. Attempt Factoring:
We look for two numbers that multiply to 4 and add to 4.
\(x^2 + 4x + 4 = (x+2)(x+2) = (x+2)^2\).
2. Solve for Roots:
\(x + 2 = 0 \implies x = -2\).
3. Count Solutions:
There is only one distinct value for \(x\).
Result: 1 Real Solution

✅ Practix Way (Optimal)

**Step 1:** Calculate the Discriminant \(D\).
\(b^2 - 4ac = 4^2 - 4(1)(4) = 16 - 16 = 0\).
**Step 2:** Interpret result.
\(D = 0 \implies \text{Exactly 1 real solution}\).
**Result: 1**
Essential for equations that don't factor easily.

🎓 Theoretical Proof: The Discriminant

Step 1: The Standard Form
Every quadratic equation of degree 2 can be expressed as:

\( ax^2 + bx + c = 0 \)

Step 2: The Quadratic Formula
The solutions (roots) of this equation are given by the quadratic formula:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Step 3: Define the Discriminant
The term under the square root, \( b^2 - 4ac \), is called the discriminant, denoted by \( D \).

\( D = b^2 - 4ac \)

Step 4: Interpret the Discriminant
The value of \( D \) determines the nature and number of real solutions:

  • If \( D > 0 \): The square root \( \sqrt{D} \) is a real, non-zero number. This leads to two distinct real solutions: \( x_1 = \frac{-b + \sqrt{D}}{2a} \) and \( x_2 = \frac{-b - \sqrt{D}}{2a} \).
  • If \( D = 0 \): The square root \( \sqrt{D} \) is zero. This leads to exactly one real solution (a repeated root): \( x = \frac{-b}{2a} \).
  • If \( D < 0 \): The square root \( \sqrt{D} \) is an imaginary number (e.g., \( \sqrt{-4} = 2i \)). This means there are no real solutions, but two complex conjugate solutions.
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