Quadratic Equations - Ninth Grade Math
Introduction to Quadratic Equations
Quadratic Equation: An equation of the form $ax^2 + bx + c = 0$ where $a \neq 0$
Standard Form: $ax^2 + bx + c = 0$
Solution/Root/Zero: Values of $x$ that make the equation true
A quadratic equation can have:
• Two distinct real solutions
• One real solution (repeated root)
• No real solutions (two complex solutions)
Standard Form: $ax^2 + bx + c = 0$
Solution/Root/Zero: Values of $x$ that make the equation true
A quadratic equation can have:
• Two distinct real solutions
• One real solution (repeated root)
• No real solutions (two complex solutions)
Methods to Solve Quadratic Equations:
1. Square Root Method
2. Factoring (using Zero Product Property)
3. Completing the Square
4. Quadratic Formula
Choose method based on the equation form!
1. Square Root Method
2. Factoring (using Zero Product Property)
3. Completing the Square
4. Quadratic Formula
Choose method based on the equation form!
1. Solve Using Square Roots
When to Use: Equations in the form $x^2 = k$ or $(x - h)^2 = k$
Square Root Property:
If $x^2 = k$, then:
$$x = \pm\sqrt{k}$$
General Form:
If $(x - h)^2 = k$, then:
$$x - h = \pm\sqrt{k}$$
$$x = h \pm\sqrt{k}$$
Remember: Always include both positive and negative square roots (±)
If $x^2 = k$, then:
$$x = \pm\sqrt{k}$$
General Form:
If $(x - h)^2 = k$, then:
$$x - h = \pm\sqrt{k}$$
$$x = h \pm\sqrt{k}$$
Remember: Always include both positive and negative square roots (±)
Steps:
Step 1: Isolate the squared term
Step 2: Take square root of both sides
Step 3: Include ± symbol
Step 4: Solve for $x$
Step 5: Simplify if needed
Step 1: Isolate the squared term
Step 2: Take square root of both sides
Step 3: Include ± symbol
Step 4: Solve for $x$
Step 5: Simplify if needed
Example 1: Solve $x^2 = 25$
$x = \pm\sqrt{25}$
$x = \pm 5$
Solutions: $x = 5$ or $x = -5$
$x = \pm\sqrt{25}$
$x = \pm 5$
Solutions: $x = 5$ or $x = -5$
Example 2: Solve $3x^2 = 48$
Step 1: Isolate $x^2$
$x^2 = 16$
Step 2: Take square root
$x = \pm\sqrt{16} = \pm 4$
Solutions: $x = 4$ or $x = -4$
Step 1: Isolate $x^2$
$x^2 = 16$
Step 2: Take square root
$x = \pm\sqrt{16} = \pm 4$
Solutions: $x = 4$ or $x = -4$
Example 3: Solve $(x - 3)^2 = 16$
$x - 3 = \pm\sqrt{16}$
$x - 3 = \pm 4$
$x = 3 + 4 = 7$ or $x = 3 - 4 = -1$
Solutions: $x = 7$ or $x = -1$
$x - 3 = \pm\sqrt{16}$
$x - 3 = \pm 4$
$x = 3 + 4 = 7$ or $x = 3 - 4 = -1$
Solutions: $x = 7$ or $x = -1$
Example 4: Solve $2(x + 1)^2 - 18 = 0$
Isolate squared term:
$2(x + 1)^2 = 18$
$(x + 1)^2 = 9$
$x + 1 = \pm 3$
$x = -1 \pm 3$
Solutions: $x = 2$ or $x = -4$
Isolate squared term:
$2(x + 1)^2 = 18$
$(x + 1)^2 = 9$
$x + 1 = \pm 3$
$x = -1 \pm 3$
Solutions: $x = 2$ or $x = -4$
Special Cases:
• If $x^2 = 0$, then $x = 0$ (one solution)
• If $x^2 = -k$ (negative), no real solutions
• Always simplify radicals: $\sqrt{18} = 3\sqrt{2}$
• If $x^2 = 0$, then $x = 0$ (one solution)
• If $x^2 = -k$ (negative), no real solutions
• Always simplify radicals: $\sqrt{18} = 3\sqrt{2}$
2-3. Zero Product Property and Solve by Factoring
Zero Product Property: If the product of two factors equals zero, then at least one factor must equal zero
$$\text{If } a \times b = 0, \text{ then } a = 0 \text{ or } b = 0$$
$$\text{If } a \times b = 0, \text{ then } a = 0 \text{ or } b = 0$$
Zero Product Property Formula:
If $(x - p)(x - q) = 0$, then:
$$x - p = 0 \text{ or } x - q = 0$$
$$x = p \text{ or } x = q$$
If $(x - p)(x - q) = 0$, then:
$$x - p = 0 \text{ or } x - q = 0$$
$$x = p \text{ or } x = q$$
Steps to Solve by Factoring:
Step 1: Write equation in standard form: $ax^2 + bx + c = 0$
Step 2: Factor the left side completely
Step 3: Set each factor equal to zero
Step 4: Solve each equation
Step 5: Check solutions (optional)
Step 1: Write equation in standard form: $ax^2 + bx + c = 0$
Step 2: Factor the left side completely
Step 3: Set each factor equal to zero
Step 4: Solve each equation
Step 5: Check solutions (optional)
Example 1: Solve $(x - 4)(x + 2) = 0$
Apply Zero Product Property:
$x - 4 = 0$ or $x + 2 = 0$
$x = 4$ or $x = -2$
Solutions: $x = 4$ or $x = -2$
Apply Zero Product Property:
$x - 4 = 0$ or $x + 2 = 0$
$x = 4$ or $x = -2$
Solutions: $x = 4$ or $x = -2$
Example 2: Solve $x^2 + 5x + 6 = 0$
Step 1: Already in standard form
Step 2: Factor
$(x + 2)(x + 3) = 0$
Step 3: Set each factor to zero
$x + 2 = 0$ or $x + 3 = 0$
Step 4: Solve
$x = -2$ or $x = -3$
Solutions: $x = -2$ or $x = -3$
Step 1: Already in standard form
Step 2: Factor
$(x + 2)(x + 3) = 0$
Step 3: Set each factor to zero
$x + 2 = 0$ or $x + 3 = 0$
Step 4: Solve
$x = -2$ or $x = -3$
Solutions: $x = -2$ or $x = -3$
Example 3: Solve $x^2 = 7x$
Step 1: Standard form
$x^2 - 7x = 0$
Step 2: Factor out GCF
$x(x - 7) = 0$
Step 3: Apply property
$x = 0$ or $x - 7 = 0$
$x = 0$ or $x = 7$
Solutions: $x = 0$ or $x = 7$
Step 1: Standard form
$x^2 - 7x = 0$
Step 2: Factor out GCF
$x(x - 7) = 0$
Step 3: Apply property
$x = 0$ or $x - 7 = 0$
$x = 0$ or $x = 7$
Solutions: $x = 0$ or $x = 7$
Example 4: Solve $2x^2 + 5x - 3 = 0$
Factor using AC method:
$(2x - 1)(x + 3) = 0$
$2x - 1 = 0$ or $x + 3 = 0$
$x = \frac{1}{2}$ or $x = -3$
Solutions: $x = \frac{1}{2}$ or $x = -3$
Factor using AC method:
$(2x - 1)(x + 3) = 0$
$2x - 1 = 0$ or $x + 3 = 0$
$x = \frac{1}{2}$ or $x = -3$
Solutions: $x = \frac{1}{2}$ or $x = -3$
Important:
• Equation MUST equal zero before applying zero product property
• Don't forget to factor out GCF first
• Check for special patterns (difference of squares, perfect squares)
• Equation MUST equal zero before applying zero product property
• Don't forget to factor out GCF first
• Check for special patterns (difference of squares, perfect squares)
4-5. Complete the Square & Solve by Completing the Square
Completing the Square: A method to rewrite a quadratic expression as a perfect square trinomial
Goal: Transform $ax^2 + bx + c$ into $a(x - h)^2 + k$
Goal: Transform $ax^2 + bx + c$ into $a(x - h)^2 + k$
Completing the Square Formula:
To complete the square for $x^2 + bx$:
Add and subtract: $\left(\frac{b}{2}\right)^2$
$$x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2$$
Key: Take half of the coefficient of $x$, then square it
To complete the square for $x^2 + bx$:
Add and subtract: $\left(\frac{b}{2}\right)^2$
$$x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2$$
Key: Take half of the coefficient of $x$, then square it
Steps to Complete the Square:
Step 1: Make coefficient of $x^2$ equal to 1 (divide all terms if needed)
Step 2: Move constant to the right side
Step 3: Take half of coefficient of $x$, square it
Step 4: Add this value to BOTH sides
Step 5: Factor the left side as a perfect square
Step 6: Solve using square root method
Step 1: Make coefficient of $x^2$ equal to 1 (divide all terms if needed)
Step 2: Move constant to the right side
Step 3: Take half of coefficient of $x$, square it
Step 4: Add this value to BOTH sides
Step 5: Factor the left side as a perfect square
Step 6: Solve using square root method
Example 1: Complete the square: $x^2 + 6x$
Half of 6: $\frac{6}{2} = 3$
Square it: $3^2 = 9$
$x^2 + 6x + 9 = (x + 3)^2$
Answer: Add 9 to complete the square
Half of 6: $\frac{6}{2} = 3$
Square it: $3^2 = 9$
$x^2 + 6x + 9 = (x + 3)^2$
Answer: Add 9 to complete the square
Example 2: Solve $x^2 + 8x - 9 = 0$ by completing the square
Step 1: Move constant
$x^2 + 8x = 9$
Step 2: Complete the square
Half of 8 is 4, $4^2 = 16$
$x^2 + 8x + 16 = 9 + 16$
Step 3: Factor
$(x + 4)^2 = 25$
Step 4: Take square root
$x + 4 = \pm 5$
$x = -4 + 5 = 1$ or $x = -4 - 5 = -9$
Solutions: $x = 1$ or $x = -9$
Step 1: Move constant
$x^2 + 8x = 9$
Step 2: Complete the square
Half of 8 is 4, $4^2 = 16$
$x^2 + 8x + 16 = 9 + 16$
Step 3: Factor
$(x + 4)^2 = 25$
Step 4: Take square root
$x + 4 = \pm 5$
$x = -4 + 5 = 1$ or $x = -4 - 5 = -9$
Solutions: $x = 1$ or $x = -9$
Example 3: Solve $x^2 - 10x + 21 = 0$
$x^2 - 10x = -21$
Half of -10 is -5, $(-5)^2 = 25$
$x^2 - 10x + 25 = -21 + 25$
$(x - 5)^2 = 4$
$x - 5 = \pm 2$
$x = 5 + 2 = 7$ or $x = 5 - 2 = 3$
Solutions: $x = 7$ or $x = 3$
$x^2 - 10x = -21$
Half of -10 is -5, $(-5)^2 = 25$
$x^2 - 10x + 25 = -21 + 25$
$(x - 5)^2 = 4$
$x - 5 = \pm 2$
$x = 5 + 2 = 7$ or $x = 5 - 2 = 3$
Solutions: $x = 7$ or $x = 3$
Example 4: Solve $2x^2 + 12x - 14 = 0$
Step 1: Divide by 2
$x^2 + 6x - 7 = 0$
$x^2 + 6x = 7$
$x^2 + 6x + 9 = 7 + 9$
$(x + 3)^2 = 16$
$x + 3 = \pm 4$
$x = -3 + 4 = 1$ or $x = -3 - 4 = -7$
Solutions: $x = 1$ or $x = -7$
Step 1: Divide by 2
$x^2 + 6x - 7 = 0$
$x^2 + 6x = 7$
$x^2 + 6x + 9 = 7 + 9$
$(x + 3)^2 = 16$
$x + 3 = \pm 4$
$x = -3 + 4 = 1$ or $x = -3 - 4 = -7$
Solutions: $x = 1$ or $x = -7$
6. Solve Using Quadratic Formula
Quadratic Formula: A formula that solves ANY quadratic equation
Works when: Factoring is difficult or impossible
Works when: Factoring is difficult or impossible
The Quadratic Formula:
For $ax^2 + bx + c = 0$ where $a \neq 0$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where:
• $a$ = coefficient of $x^2$
• $b$ = coefficient of $x$
• $c$ = constant term
• $\pm$ means two solutions: one with +, one with -
For $ax^2 + bx + c = 0$ where $a \neq 0$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where:
• $a$ = coefficient of $x^2$
• $b$ = coefficient of $x$
• $c$ = constant term
• $\pm$ means two solutions: one with +, one with -
Steps to Use Quadratic Formula:
Step 1: Write equation in standard form: $ax^2 + bx + c = 0$
Step 2: Identify values of $a$, $b$, and $c$
Step 3: Substitute into formula
Step 4: Simplify under the square root first
Step 5: Simplify the entire expression
Step 6: Write both solutions
Step 1: Write equation in standard form: $ax^2 + bx + c = 0$
Step 2: Identify values of $a$, $b$, and $c$
Step 3: Substitute into formula
Step 4: Simplify under the square root first
Step 5: Simplify the entire expression
Step 6: Write both solutions
Example 1: Solve $x^2 + 5x + 6 = 0$ using quadratic formula
Step 1: Identify $a = 1$, $b = 5$, $c = 6$
Step 2: Substitute
$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$
Step 3: Simplify
$x = \frac{-5 \pm \sqrt{25 - 24}}{2}$
$x = \frac{-5 \pm \sqrt{1}}{2}$
$x = \frac{-5 \pm 1}{2}$
$x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$ or $x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$
Solutions: $x = -2$ or $x = -3$
Step 1: Identify $a = 1$, $b = 5$, $c = 6$
Step 2: Substitute
$x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}$
Step 3: Simplify
$x = \frac{-5 \pm \sqrt{25 - 24}}{2}$
$x = \frac{-5 \pm \sqrt{1}}{2}$
$x = \frac{-5 \pm 1}{2}$
$x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$ or $x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$
Solutions: $x = -2$ or $x = -3$
Example 2: Solve $2x^2 - 7x + 3 = 0$
$a = 2$, $b = -7$, $c = 3$
$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$
$x = \frac{7 \pm \sqrt{49 - 24}}{4}$
$x = \frac{7 \pm \sqrt{25}}{4}$
$x = \frac{7 \pm 5}{4}$
$x = \frac{7 + 5}{4} = 3$ or $x = \frac{7 - 5}{4} = \frac{1}{2}$
Solutions: $x = 3$ or $x = \frac{1}{2}$
$a = 2$, $b = -7$, $c = 3$
$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$
$x = \frac{7 \pm \sqrt{49 - 24}}{4}$
$x = \frac{7 \pm \sqrt{25}}{4}$
$x = \frac{7 \pm 5}{4}$
$x = \frac{7 + 5}{4} = 3$ or $x = \frac{7 - 5}{4} = \frac{1}{2}$
Solutions: $x = 3$ or $x = \frac{1}{2}$
Example 3: Solve $x^2 - 6x + 2 = 0$
$a = 1$, $b = -6$, $c = 2$
$x = \frac{6 \pm \sqrt{36 - 8}}{2}$
$x = \frac{6 \pm \sqrt{28}}{2}$
$x = \frac{6 \pm 2\sqrt{7}}{2}$
$x = 3 \pm \sqrt{7}$
Solutions: $x = 3 + \sqrt{7}$ or $x = 3 - \sqrt{7}$
$a = 1$, $b = -6$, $c = 2$
$x = \frac{6 \pm \sqrt{36 - 8}}{2}$
$x = \frac{6 \pm \sqrt{28}}{2}$
$x = \frac{6 \pm 2\sqrt{7}}{2}$
$x = 3 \pm \sqrt{7}$
Solutions: $x = 3 + \sqrt{7}$ or $x = 3 - \sqrt{7}$
Tips:
• Be careful with negative signs!
• Simplify radicals completely
• Can use for ANY quadratic equation
• Check if you can simplify the fraction
• Be careful with negative signs!
• Simplify radicals completely
• Can use for ANY quadratic equation
• Check if you can simplify the fraction
7. Solve Quadratic Equations: Word Problems
Common Applications:
• Area problems
• Projectile motion
• Number problems
• Consecutive integers
• Geometry problems
• Area problems
• Projectile motion
• Number problems
• Consecutive integers
• Geometry problems
Steps for Word Problems:
Step 1: Read carefully and identify what you're looking for
Step 2: Define variable(s)
Step 3: Write equation from given information
Step 4: Solve using appropriate method
Step 5: Check if solutions make sense in context
Step 6: Answer the question with units
Step 1: Read carefully and identify what you're looking for
Step 2: Define variable(s)
Step 3: Write equation from given information
Step 4: Solve using appropriate method
Step 5: Check if solutions make sense in context
Step 6: Answer the question with units
Example 1: A rectangle has length that is 3 feet more than its width. If the area is 54 square feet, find the dimensions.
Step 1: Let $w$ = width
Then length = $w + 3$
Step 2: Area equation
$w(w + 3) = 54$
$w^2 + 3w = 54$
$w^2 + 3w - 54 = 0$
Step 3: Factor
$(w + 9)(w - 6) = 0$
$w = -9$ or $w = 6$
Step 4: Check context
Width cannot be negative, so $w = 6$ feet
Length = $6 + 3 = 9$ feet
Answer: Width = 6 ft, Length = 9 ft
Step 1: Let $w$ = width
Then length = $w + 3$
Step 2: Area equation
$w(w + 3) = 54$
$w^2 + 3w = 54$
$w^2 + 3w - 54 = 0$
Step 3: Factor
$(w + 9)(w - 6) = 0$
$w = -9$ or $w = 6$
Step 4: Check context
Width cannot be negative, so $w = 6$ feet
Length = $6 + 3 = 9$ feet
Answer: Width = 6 ft, Length = 9 ft
Example 2: A ball is thrown upward with initial velocity 64 ft/s from height of 80 feet. Height equation: $h(t) = -16t^2 + 64t + 80$. When does it hit the ground?
Ground means $h = 0$:
$0 = -16t^2 + 64t + 80$
$0 = -16(t^2 - 4t - 5)$
$0 = -16(t - 5)(t + 1)$
$t = 5$ or $t = -1$
Time cannot be negative
Answer: Ball hits ground at $t = 5$ seconds
Ground means $h = 0$:
$0 = -16t^2 + 64t + 80$
$0 = -16(t^2 - 4t - 5)$
$0 = -16(t - 5)(t + 1)$
$t = 5$ or $t = -1$
Time cannot be negative
Answer: Ball hits ground at $t = 5$ seconds
Example 3: The product of two consecutive positive integers is 132. Find the integers.
Let $n$ = first integer
Then $n + 1$ = second integer
$n(n + 1) = 132$
$n^2 + n = 132$
$n^2 + n - 132 = 0$
$(n + 12)(n - 11) = 0$
$n = -12$ or $n = 11$
Since we need positive integers: $n = 11$
Next integer: $11 + 1 = 12$
Answer: 11 and 12
Let $n$ = first integer
Then $n + 1$ = second integer
$n(n + 1) = 132$
$n^2 + n = 132$
$n^2 + n - 132 = 0$
$(n + 12)(n - 11) = 0$
$n = -12$ or $n = 11$
Since we need positive integers: $n = 11$
Next integer: $11 + 1 = 12$
Answer: 11 and 12
8. Using the Discriminant
Discriminant: The expression under the square root in the quadratic formula
Symbol: $\Delta$ or $D$
Purpose: Determines the number and type of solutions WITHOUT solving
Symbol: $\Delta$ or $D$
Purpose: Determines the number and type of solutions WITHOUT solving
Discriminant Formula:
For $ax^2 + bx + c = 0$:
$$\Delta = b^2 - 4ac$$
Interpretation:
• If $\Delta > 0$: Two distinct real solutions
(Parabola crosses x-axis twice)
• If $\Delta = 0$: One real solution (repeated root)
(Parabola touches x-axis at vertex)
• If $\Delta < 0$: No real solutions
(Two complex solutions; parabola doesn't cross x-axis)
For $ax^2 + bx + c = 0$:
$$\Delta = b^2 - 4ac$$
Interpretation:
• If $\Delta > 0$: Two distinct real solutions
(Parabola crosses x-axis twice)
• If $\Delta = 0$: One real solution (repeated root)
(Parabola touches x-axis at vertex)
• If $\Delta < 0$: No real solutions
(Two complex solutions; parabola doesn't cross x-axis)
Example 1: Find discriminant and number of solutions: $x^2 + 5x + 6 = 0$
$a = 1$, $b = 5$, $c = 6$
$\Delta = 5^2 - 4(1)(6)$
$\Delta = 25 - 24 = 1$
Since $\Delta > 0$: Two distinct real solutions
Answer: Two real solutions
$a = 1$, $b = 5$, $c = 6$
$\Delta = 5^2 - 4(1)(6)$
$\Delta = 25 - 24 = 1$
Since $\Delta > 0$: Two distinct real solutions
Answer: Two real solutions
Example 2: $x^2 - 6x + 9 = 0$
$\Delta = (-6)^2 - 4(1)(9)$
$\Delta = 36 - 36 = 0$
Since $\Delta = 0$: One real solution
(This is a perfect square: $(x - 3)^2 = 0$, so $x = 3$)
Answer: One real solution
$\Delta = (-6)^2 - 4(1)(9)$
$\Delta = 36 - 36 = 0$
Since $\Delta = 0$: One real solution
(This is a perfect square: $(x - 3)^2 = 0$, so $x = 3$)
Answer: One real solution
Example 3: $2x^2 + 3x + 5 = 0$
$\Delta = 3^2 - 4(2)(5)$
$\Delta = 9 - 40 = -31$
Since $\Delta < 0$: No real solutions
Answer: No real solutions
$\Delta = 3^2 - 4(2)(5)$
$\Delta = 9 - 40 = -31$
Since $\Delta < 0$: No real solutions
Answer: No real solutions
Example 4: For what value of $k$ does $x^2 + kx + 16 = 0$ have exactly one solution?
One solution means $\Delta = 0$:
$k^2 - 4(1)(16) = 0$
$k^2 - 64 = 0$
$k^2 = 64$
$k = \pm 8$
Answer: $k = 8$ or $k = -8$
One solution means $\Delta = 0$:
$k^2 - 4(1)(16) = 0$
$k^2 - 64 = 0$
$k^2 = 64$
$k = \pm 8$
Answer: $k = 8$ or $k = -8$
Perfect Square Discriminant:
If $\Delta$ is a perfect square (and positive), solutions are rational
If $\Delta$ is positive but not perfect square, solutions are irrational
If $\Delta$ is a perfect square (and positive), solutions are rational
If $\Delta$ is positive but not perfect square, solutions are irrational
9-10. Systems of Linear and Quadratic Equations
System: Two equations with two variables
Solution: Point(s) $(x, y)$ that satisfy BOTH equations
Possible Solutions:
• Two points (line intersects parabola twice)
• One point (line is tangent to parabola)
• No points (line doesn't intersect parabola)
Solution: Point(s) $(x, y)$ that satisfy BOTH equations
Possible Solutions:
• Two points (line intersects parabola twice)
• One point (line is tangent to parabola)
• No points (line doesn't intersect parabola)
Graphing Method
Steps to Solve by Graphing:
Step 1: Graph the linear equation
Step 2: Graph the quadratic equation
Step 3: Find intersection point(s)
Step 4: Write solution(s) as ordered pairs
Step 1: Graph the linear equation
Step 2: Graph the quadratic equation
Step 3: Find intersection point(s)
Step 4: Write solution(s) as ordered pairs
Example 1: Solve graphically:
$y = x^2 - 4$
$y = x + 2$
Graph both equations
Parabola opens up with vertex at $(0, -4)$
Line has slope 1, y-intercept 2
Intersection points from graph:
$(-1, 1)$ and $(3, 5)$
Solutions: $(-1, 1)$ and $(3, 5)$
$y = x^2 - 4$
$y = x + 2$
Graph both equations
Parabola opens up with vertex at $(0, -4)$
Line has slope 1, y-intercept 2
Intersection points from graph:
$(-1, 1)$ and $(3, 5)$
Solutions: $(-1, 1)$ and $(3, 5)$
Algebraic Method (Substitution)
Steps to Solve Algebraically:
Step 1: Solve linear equation for one variable
Step 2: Substitute into quadratic equation
Step 3: Solve resulting quadratic equation
Step 4: Substitute back to find other coordinate
Step 5: Write solutions as ordered pairs
Step 1: Solve linear equation for one variable
Step 2: Substitute into quadratic equation
Step 3: Solve resulting quadratic equation
Step 4: Substitute back to find other coordinate
Step 5: Write solutions as ordered pairs
Example 2: Solve:
$y = x^2 - 4$
$y = x + 2$
Step 1: Since both equal $y$:
$x^2 - 4 = x + 2$
Step 2: Standard form
$x^2 - x - 6 = 0$
Step 3: Factor
$(x - 3)(x + 2) = 0$
$x = 3$ or $x = -2$
Step 4: Find $y$ values
When $x = 3$: $y = 3 + 2 = 5$
When $x = -2$: $y = -2 + 2 = 0$
Solutions: $(3, 5)$ and $(-2, 0)$
$y = x^2 - 4$
$y = x + 2$
Step 1: Since both equal $y$:
$x^2 - 4 = x + 2$
Step 2: Standard form
$x^2 - x - 6 = 0$
Step 3: Factor
$(x - 3)(x + 2) = 0$
$x = 3$ or $x = -2$
Step 4: Find $y$ values
When $x = 3$: $y = 3 + 2 = 5$
When $x = -2$: $y = -2 + 2 = 0$
Solutions: $(3, 5)$ and $(-2, 0)$
Example 3: Solve:
$y = x^2 + 2x - 3$
$y = 2x + 1$
Set equal:
$x^2 + 2x - 3 = 2x + 1$
$x^2 - 4 = 0$
$x^2 = 4$
$x = \pm 2$
Find $y$:
When $x = 2$: $y = 2(2) + 1 = 5$ → $(2, 5)$
When $x = -2$: $y = 2(-2) + 1 = -3$ → $(-2, -3)$
Solutions: $(2, 5)$ and $(-2, -3)$
$y = x^2 + 2x - 3$
$y = 2x + 1$
Set equal:
$x^2 + 2x - 3 = 2x + 1$
$x^2 - 4 = 0$
$x^2 = 4$
$x = \pm 2$
Find $y$:
When $x = 2$: $y = 2(2) + 1 = 5$ → $(2, 5)$
When $x = -2$: $y = 2(-2) + 1 = -3$ → $(-2, -3)$
Solutions: $(2, 5)$ and $(-2, -3)$
Example 4: Solve:
$x^2 + y^2 = 25$
$y = x + 1$
Substitute $y = x + 1$:
$x^2 + (x + 1)^2 = 25$
$x^2 + x^2 + 2x + 1 = 25$
$2x^2 + 2x - 24 = 0$
$x^2 + x - 12 = 0$
$(x + 4)(x - 3) = 0$
$x = -4$ or $x = 3$
Find $y$:
When $x = -4$: $y = -4 + 1 = -3$ → $(-4, -3)$
When $x = 3$: $y = 3 + 1 = 4$ → $(3, 4)$
Solutions: $(-4, -3)$ and $(3, 4)$
$x^2 + y^2 = 25$
$y = x + 1$
Substitute $y = x + 1$:
$x^2 + (x + 1)^2 = 25$
$x^2 + x^2 + 2x + 1 = 25$
$2x^2 + 2x - 24 = 0$
$x^2 + x - 12 = 0$
$(x + 4)(x - 3) = 0$
$x = -4$ or $x = 3$
Find $y$:
When $x = -4$: $y = -4 + 1 = -3$ → $(-4, -3)$
When $x = 3$: $y = 3 + 1 = 4$ → $(3, 4)$
Solutions: $(-4, -3)$ and $(3, 4)$
Summary: Methods to Solve Quadratic Equations
| Method | Best Used When | Formula/Steps | Example |
|---|---|---|---|
| Square Roots | Equation in form $x^2 = k$ or $(x-h)^2 = k$ | $x = \pm\sqrt{k}$ | $x^2 = 16$ → $x = \pm 4$ |
| Factoring | Can factor easily | Factor, set each = 0 | $x^2 + 5x + 6 = 0$ → $(x+2)(x+3) = 0$ |
| Completing Square | $a = 1$ or converting forms | Add $\left(\frac{b}{2}\right)^2$ to both sides | $x^2 + 6x = 7$ → $(x+3)^2 = 16$ |
| Quadratic Formula | Any quadratic; when factoring fails | $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ | Works for ALL quadratics |
Discriminant Quick Reference
| Discriminant Value | Number of Solutions | Type of Solutions | Graph |
|---|---|---|---|
| $\Delta > 0$ (positive) | 2 distinct solutions | Real and unequal | Crosses x-axis twice |
| $\Delta = 0$ (zero) | 1 solution | Real and equal (repeated) | Touches x-axis once |
| $\Delta < 0$ (negative) | 0 real solutions | 2 complex (imaginary) | Doesn't cross x-axis |
Decision Flowchart: Which Method?
Choose Your Method:
1. Is the equation in form $x^2 = k$?
→ YES: Use square root method
→ NO: Continue
2. Can you factor easily?
→ YES: Factor and use zero product property
→ NO: Continue
3. Do you need exact or approximate answer?
→ EXACT: Use quadratic formula
→ APPROXIMATE: Graph or use calculator
4. Need to convert to vertex form?
→ YES: Complete the square
When in doubt: QUADRATIC FORMULA works for everything!
1. Is the equation in form $x^2 = k$?
→ YES: Use square root method
→ NO: Continue
2. Can you factor easily?
→ YES: Factor and use zero product property
→ NO: Continue
3. Do you need exact or approximate answer?
→ EXACT: Use quadratic formula
→ APPROXIMATE: Graph or use calculator
4. Need to convert to vertex form?
→ YES: Complete the square
When in doubt: QUADRATIC FORMULA works for everything!
Success Tips for Solving Quadratic Equations:
✓ Always write in standard form first: $ax^2 + bx + c = 0$
✓ Remember ± when using square roots
✓ Equation must equal zero before factoring
✓ Memorize quadratic formula - it always works!
✓ Use discriminant to check number of solutions
✓ Simplify radicals completely
✓ Check solutions in original equation
✓ In word problems, verify solutions make sense
✓ Practice all methods - each has advantages
✓ For systems: substitution is usually easiest
✓ Always write in standard form first: $ax^2 + bx + c = 0$
✓ Remember ± when using square roots
✓ Equation must equal zero before factoring
✓ Memorize quadratic formula - it always works!
✓ Use discriminant to check number of solutions
✓ Simplify radicals completely
✓ Check solutions in original equation
✓ In word problems, verify solutions make sense
✓ Practice all methods - each has advantages
✓ For systems: substitution is usually easiest