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Quadratic Roots
Standard Form and the Parabola
A quadratic equation is a second-degree polynomial in standard form $ax^2 + bx + c = 0$, representing a parabola.
$$ax^2 + bx + c = 0$$
The standard form where 'a' determines the curve's opening direction.
$a$=Leading coefficient (quadratic term)(constant)
$b$=Linear coefficient(constant)
$c$=Constant term (y-intercept)(constant)
$a > 0$
→Parabola opens upwards (U-shape)
$a < 0$
→Parabola opens downwards (inverted U)
Geometric Meaning: The roots are the x-values where the curve intersects the horizontal axis.
Roots vs Intercepts: In mathematics, the terms roots, zeros, and x-intercepts are often used interchangeably for these values.
Predicting with the Discriminant
The discriminant ($D$) determines the nature of the roots without solving the entire equation.
$$D = b^2 - 4ac$$
The part of the quadratic formula under the radical.
$D$=Discriminant(unitless)
$D > 0$
→Two real and distinct roots
$D = 0$
→One real and equal root (repeated)
$D < 0$
→Two complex / imaginary roots
Perfect Squares: If $a, b, c$ are rational and $D$ is a perfect square, the roots are rational.
Irrational Roots: If $D > 0$ but not a perfect square (with rational $a, b, c$), the roots are irrational and come in conjugate pairs like $p \pm \sqrt{q}$.
Complex Conjugates: When $D < 0$, complex roots always appear in conjugate pairs $p \pm qi$.
Negative Reality: A negative $D$ implies the parabola never touches the x-axis, existing entirely above or below it.
The Quadratic Formula
The Quadratic Formula provides the exact values for the roots using the coefficients $a, b,$ and $c$.
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Derives solutions for any quadratic equation.
$\pm$=Plus-minus sign(indicates two solutions)
$\sqrt{...}$=[radical] expression(operation)
$\sqrt{D} = 0$
→$x = -b / 2a$ (vertex position)
Sign Safety: Always keep the sign of $b$ consistent; $-b$ means 'opposite of $b$'.
Calculation Order: Always compute the discriminant $D$ first to verify if real roots exist before proceeding.
Vieta's Formulas
Vieta's formulas relate the sum and product of the roots directly to the coefficients.
$$\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}$$
Bypasses finding individual roots to find their collective properties.
$\alpha, \beta$=The two roots of the equation(x-values)
$\alpha + \beta$=Sum of roots(unitless)
$\alpha\beta$=Product of roots(unitless)
$a = 1$
→monic quadratic: Sum = -b, Product = c
Monic Advantage: In a monic equation ($x^2 + px + q = 0$), the sum is simply the negative coefficient of $x$.
Constructing Equations: Given roots $\alpha, \beta$, the equation is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.
Symmetric Expressions: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ lets you compute higher expressions without finding individual roots.
Difference of Roots: $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \frac{\sqrt{D}}{|a|}$.
Symmetry: These formulas hold true even for complex roots.
Cube Roots of Unity
The cube roots of unity are the three solutions to $x^3 = 1$, consisting of one real root and two complex conjugate roots.
$$1, \quad \omega = \frac{-1 + \sqrt{3}i}{2}, \quad \omega^2 = \frac{-1 - \sqrt{3}i}{2}$$
Found by solving $x^3 - 1 = 0 \implies (x-1)(x^2+x+1) = 0$.
$\omega$=Complex cube root of unity (omega)
$\omega^2$=Second complex cube root
$n$ is a multiple of 3
→$\omega^n = 1$
$n \equiv 1 \pmod{3}$
→$\omega^n = \omega$
$n \equiv 2 \pmod{3}$
→$\omega^n = \omega^2$
Sum Property: $1 + \omega + \omega^2 = 0$, equivalently $\omega + \omega^2 = -1$.
Product Property: $\omega^3 = 1$ (product of all three roots is unity).
Conjugate Pair: $\omega$ and $\omega^2$ are complex conjugates of each other.
Exponent Reduction: For any integer $n$, divide exponent by 3 and use the remainder: $\omega^{11} = \omega^{9} \cdot \omega^{2} = (\omega^3)^3 \cdot \omega^2 = \omega^2$.
Factorization: $x^3 + y^3 = (x+y)(x+\omega y)(x+\omega^2 y)$ and $x^3 - y^3 = (x-y)(x-\omega y)(x-\omega^2 y)$.
Fourth Roots of Unity
The fourth roots of unity are the four solutions to $x^4 = 1$, consisting of two real and two imaginary values.
$$+1, \quad -1, \quad +i, \quad -i$$
Found by solving $x^4 - 1 = 0 \implies (x^2-1)(x^2+1) = 0$.
$\pm 1$=Real fourth roots
$\pm i$=Imaginary fourth roots
Sum Property: $1 + (-1) + i + (-i) = 0$ (sum of all fourth roots is zero).
Product Property: $1 \times (-1) \times i \times (-i) = -1$ (product of all fourth roots is $-1$, unlike cube roots).
Real Roots: $+1$ and $-1$ are additive inverses.
Imaginary Roots: $+i$ and $-i$ are complex conjugates.
Finding $n$th Roots: For $\sqrt[4]{a}$, multiply each fourth root of unity by $\sqrt[4]{|a|}$. For example, $\sqrt[4]{16} = 2 \times \{1, -1, i, -i\} = \{2, -2, 2i, -2i\}$.
Application Strategy
Worked Examples
Problem Statement
Solve the quadratic equation $x^2 - 5x + 6 = 0$ using the quadratic formula.
Standard FormQuadratic Formula
Strategic Toolkit
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Step 1
Identify Coefficients
Compare $x^2 - 5x + 6 = 0$ with $ax^2 + bx + c = 0$.
$a = 1, b = -5, c = 6$
Step 2
Calculate Discriminant
Substitute values into $D = b^2 - 4ac$.
$D = (-5)^2 - 4(1)(6) = 25 - 24 = 1$
Step 3
Apply Formula
Substitute $a, b,$ and $D$ into the quadratic formula.
$x = \frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}$
Step 4
Solve for Roots
Calculate both the plus and minus scenarios.
$x_1 = \frac{6}{2} = 3, x_2 = \frac{4}{2} = 2$
Problem Statement
For what values of $m$ does $(m+1)x^2 + 2(m+3)x + 2m+3 = 0$ have equal roots?
DiscriminantNature of RootsParameter Problems
Strategic Toolkit
$D = b^2 - 4ac = 0$
Step 1
Identify Coefficients
Map to $ax^2 + bx + c = 0$.
$a = m+1, \quad b = 2(m+3), \quad c = 2m+3$
Step 2
Compute Discriminant
Expand $D = b^2 - 4ac$.
$D = 4(m+3)^2 - 4(m+1)(2m+3) = 4(m^2+6m+9) - 4(2m^2+5m+3)$
Step 3
Simplify
Combine like terms.
$D = 4m^2+24m+36-8m^2-20m-12 = -4m^2+4m+24$
Step 4
Set D = 0
For equal roots, set $D = 0$ and divide through by $-4$:
$-4m^2 + 4m + 24 = 0 \implies m^2 - m - 6 = 0$
Step 5
Factor and Solve
Factor the quadratic in $m$:
$(m-3)(m+2) = 0 \implies m = 3 \text{ or } m = -2$
Problem Statement
If $\alpha, \beta$ are roots of $2x^2 - 6x + 3 = 0$, find $\alpha^2 + \beta^2$.
Vieta's FormulasSymmetric Expressions
Strategic Toolkit
$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$
Step 1
Apply Vieta's Formulas
With $a=2, b=-6, c=3$, compute sum and product.
$\alpha + \beta = \frac{-(-6)}{2} = 3, \quad \alpha\beta = \frac{3}{2}$
Step 2
Use Identity
Apply $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
$\alpha^2 + \beta^2 = (3)^2 - 2\left(\frac{3}{2}\right) = 9 - 3 = 6$
Problem Statement
Evaluate $(1 + \omega - \omega^2)^8$ where $\omega$ is a complex cube root of unity.
Cube Roots of UnityProperties of ω
Strategic Toolkit
$1 + \omega + \omega^2 = 0, \quad \omega^3 = 1$
Step 1
Simplify Using Sum Property
Since $1 + \omega + \omega^2 = 0$, we get $1 + \omega = -\omega^2$.
$1 + \omega - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$
Step 2
Raise to Power 8
Compute $(-2\omega^2)^8$.
$(-2)^8 \cdot (\omega^2)^8 = 256 \cdot \omega^{16}$
Step 3
Reduce Exponent
Divide 16 by 3: $16 = 3 \times 5 + 1$, so $\omega^{16} = \omega$.
$256\omega^{16} = 256 \cdot \omega^{15} \cdot \omega = 256 \cdot (\omega^3)^5 \cdot \omega = 256\omega$
Memorize formulas
Quadratic Formula
The ultimate 'X-finder' for any quadratic equation.
Formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Discriminant (D)
The nature-checker. Positive = 2, Zero = 1, Negative = Complex.
Formula
$D = b^2 - 4ac$
Sum of Roots
Negative 'b' divided by 'a'. Remember the sign flip!
Formula
$\alpha + \beta = -\frac{b}{a}$
Product of Roots
The constant 'c' divided by 'a'. No sign flip here.
Formula
$\alpha\beta = \frac{c}{a}$
Sum of Squares of Roots
Express α² + β² using sum and product — avoids solving for individual roots.
Formula
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
Cube Roots of Unity — Sum
The three cube roots of unity always sum to zero.
Formula
$1 + \omega + \omega^2 = 0$
Cube Roots of Unity — Product
The cube of omega cycles back to 1.
Formula
$\omega^3 = 1$
Review
1
For the equation $2x^2 - 8x + k = 0$ to have real and distinct roots, what must be the range of $k$?
Ans
$k < 8$
Review
Tap to reveal
Use the discriminant condition
A quadratic has real and distinct roots only when its discriminant is positive. Here, $D = b^2 - 4ac = (-8)^2 - 4(2)(k) = 64 - 8k$.
$$64 - 8k > 0 \;\Rightarrow\; k < 8$$
2
In the equation $x^2 + px + q = 0$, if the sum of roots is $5$ and the product of roots is $6$, what are the values of $p$ and $q$?
Ans
$p = -5, q = 6$
Review
Tap to reveal
Apply Vieta's formulas
For $x^2 + px + q = 0$, the sum of roots is $-p$ and the product of roots is $q$. Match these with the values given in the question.
$$-p = 5 \Rightarrow p = -5, \qquad q = 6$$
3
Which of the following conditions ensures that a quadratic equation $ax^2 + bx + c = 0$ has rational roots, given $a, b,$ and $c$ are rational?
Ans
$b^2 - 4ac$ is a non-negative perfect square
Review
Tap to reveal
Require a rational square root of the discriminant
The quadratic formula gives roots involving $\sqrt{D}$. If $a$, $b$, and $c$ are rational, the roots are rational exactly when $D = b^2 - 4ac$ is a non-negative perfect square.
$$D \ge 0 \text{ and } \sqrt{D} \in \mathbb{Q}$$
4
If $\alpha$ and $\beta$ are roots of $3x^2 - 5x + 1 = 0$, what is the value of $\alpha^3 + \beta^3$?
Ans
$\frac{80}{27}$
Review
Tap to reveal
Use the sum-product identity for cubes
From Vieta's formulas, $\alpha + \beta = \frac{5}{3}$ and $\alpha\beta = \frac{1}{3}$. Then use $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$.
$$\alpha^3+\beta^3 = \left(\frac{5}{3}\right)^3 - 3\left(\frac{1}{3}\right)\left(\frac{5}{3}\right) = \frac{125}{27} - \frac{45}{27} = \frac{80}{27}$$
5
For $x^2 - 2kx + (k^2 - k) = 0$, the values of $k$ for which the sum of the roots equals the product of the roots are:
Ans
$k = 0$ or $k = 3$
Review
Tap to reveal
Set sum equal to product
For this quadratic, the sum of roots is $2k$ and the product is $k^2-k$. Equate them and solve the resulting quadratic in $k$.
$$2k = k^2 - k \Rightarrow k^2 - 3k = 0 \Rightarrow k(k-3)=0$$
6
The value of $\omega^{28} + \omega^{29} + 1$, where $\omega$ is a complex cube root of unity, is:
Ans
$0$
Review
Tap to reveal
Reduce powers modulo 3
Since $\omega^3 = 1$, powers of $\omega$ repeat every 3. Reduce each exponent to get $\omega^{28} = \omega$ and $\omega^{29} = \omega^2$, then use the cube-root identity.
$$\omega^{28} + \omega^{29} + 1 = \omega + \omega^2 + 1 = 0$$
7
If $\omega$ is a complex cube root of unity, then $(1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^8)$ equals:
Ans
$1$
Review
Tap to reveal
Reduce each power of $\omega$
Because $\omega^3 = 1$, we have $\omega^4 = \omega$ and $\omega^8 = \omega^2$. The product becomes $(1+\omega)^2(1+\omega^2)^2$, and $ (1+\omega)(1+\omega^2) = 1$.
$$(1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^8) = \big((1+\omega)(1+\omega^2)\big)^2 = 1$$
8
The product of the four fourth roots of unity is:
Ans
$-1$
Review
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List the fourth roots directly
The fourth roots of unity are the solutions of $z^4 = 1$: $1, -1, i,$ and $-i$. Multiply them in any order.
$$1 \cdot (-1) \cdot i \cdot (-i) = -1$$
9
If $\alpha$ and $\beta$ are roots of $x^2 - 7x + 10 = 0$, the equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:
Ans
$10x^2 - 7x + 1 = 0$
Review
Tap to reveal
Invert the roots and re-form the quadratic
If the original roots have sum $\alpha+\beta$ and product $\alpha\beta$, then the reciprocals have sum $\frac{\alpha+\beta}{\alpha\beta}$ and product $\frac{1}{\alpha\beta}$. Use the given quadratic to compute those values.
$$\text{New roots: sum }= \frac{7}{10},\; \text{product }= \frac{1}{10} \Rightarrow x^2 - \frac{7}{10}x + \frac{1}{10}=0$$
10
The roots of $(p+q)x^2 - px - q = 0$ are always:
Ans
Rational
Review
Tap to reveal
Check the discriminant
For $(p+q)x^2 - px - q = 0$, the discriminant simplifies to a perfect square for every $p$ and $q$. A perfect-square discriminant guarantees rational roots.
$$D = (-p)^2 - 4(p+q)(-q) = p^2 + 4pq + 4q^2 = (p+2q)^2$$
11
If one root of $x^2 + px + 12 = 0$ is $4$, then $p$ equals:
Ans
$-7$
Review
Tap to reveal
Use the given root to find the other one
Since the product of the roots is $12$, the other root must be $12/4 = 3$. Then the sum of the roots is $7$, which equals $-p$.
$$4 + 3 = 7 = -p \Rightarrow p = -7$$
12
The three cube roots of $-27$ are:
Ans
$-3, \quad -3\omega, \quad -3\omega^2$
Review
Tap to reveal
Multiply the real cube root by cube roots of unity
The real cube root of $-27$ is $-3$. All cube roots are obtained by multiplying this value by the three cube roots of unity: $1$, $\omega$, and $\omega^2$.
$$-3, \quad -3\omega, \quad -3\omega^2$$
Question 1 of 12
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1
For the equation $2x^2 - 8x + k = 0$ to have real and distinct roots, what must be the range of $k$?