Master Roots of a Quadratic Equation Effortlessly

Building on your understanding of quadratic equations, in this article, we’re diving deep into their most important feature: the roots of a quadratic equation. Remember those solutions we find for quadratic equations? They’re not just numbers – they tell us a lot about the equation itself and the curves they represent. Here, we’ll explore what these ‘roots’ truly mean, how they are determined, and how the discriminant becomes our key to understanding their nature. We’ll also uncover fascinating connections, like the relationships between a quadratic’s coefficients and its roots, and even touch upon topics like cube roots of unity. Get ready to unlock a deeper understanding of roots of a quadratic equation.

Nature of the Roots of a Quadratic Equation

Before we learn about the roots of a quadratic equation we must have a clear concept of a quadratic equation.

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How to Calculate the Discriminant

To find the discriminant, identify the coefficients a, b, and c from the quadratic equation and substitute them into the formula.

Example 1:
Find the discriminant of
  2x² – 4x + 3 = 0

1. Compare with ax² + bx + c = 0:   a = 2, b = –4, c = 3
2. Substitute into the formula: D = (-4)² − 4 × 2 × 3 = 16 − 24 = −8

So, D = –8.

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Example 2:
Find the discriminant of
  x² + 6x + 9 = 0

1. Here, a = 1, b = 6, c = 9.
2. Calculate: D = 6² − 4 × 1 × 9 = 36 − 36 = 0

Thus, D = 0.

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What Does the Discriminant Tell Us?

• If D > 0: The equation has two different (distinct) real roots.
• If D = 0: The equation has two real roots that are equal (one unique real root repeated).
• If D < 0: The equation has no real roots; instead, the roots are complex (or imaginary).

Verifying with the Quadratic Formula

For instance, using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

• Example 1 (continued):
  For 2x² – 4x + 3 = 0, we found D = –8. Using the quadratic formula:
  \(x = \frac{4 \pm \sqrt{-8}}{4}\)
  Since √(–8) is complex, there are no real roots.
• Example 2 (continued): For x² + 6x + 9 = 0, with D = 0:
  \(x = \frac{-6 \pm \sqrt{0}}{2}= -3\)
  There is one repeated real root, x = –3.

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Using the Discriminant to Find Unknowns

Sometimes an equation includes an unknown parameter, and a condition on the roots tells us its value.

Example 4:
Find k so that

  2x² + kx + 8 = 0

has real and equal roots (i.e. D = 0).

1. Here, a = 2, b = k, c = 8.
2. Set D = 0:
   k² − 4×2×8 = k² − 64 = 0

3. Solve: k² = 64 ⇒ k = ±8

Thus, k = 8 or k = –8.

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Cube Roots of Unity and Their Properties

What Are Cube Roots of Unity?

“Unity” means 1. The cube roots of unity are the numbers that satisfy the equation x³ = 1 or x³ -1= 0

Using the difference of cubes factorization
(a³ – b³ = (a – b)(a² + ab + b²)) with a = x and b = 1,
we have:
x³ − 1 = (x − 1)(x² + x + 1) = 0

Thus, either:

• x – 1 = 0 ⟹ x = 1, or
• x² + x + 1 = 0

Solving x² + x + 1 = 0 by the quadratic formula:

\( x = \frac{-1 \pm \sqrt{1 – 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \)

So, the three cube roots of unity are:

  1, ω = (–1 + i√3)/2, ω² = (–1 – i√3)/2

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Recognizing ω and ω²

The two complex roots are labeled ω and ω². You can verify, with a bit of algebra, that:

\( \omega^2 = \frac{-1 – i\sqrt{3}}{2} \)

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Important Properties

• Sum of the cube roots:

  1 + ω + ω² = 0

• Product of the cube roots:

  1 × ω × ω² = ω³ = 1

Additional relations include:

  ω + ω² = –1, 1 + ω = –ω², and 1 + ω² = –ω

Because ω³ = 1, the powers of ω repeat every 3 steps:
ω, ω², 1, ω, ω², 1, … etc.

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Using Cube Roots of Unity in Problems

Example 5:
Find the value of

  (1 + ω – ω²)³

Step 1: Since 1 + ω + ω² = 0, we have
  1 + ω = –ω².

Step 2: Substitute into the expression:

  1 + ω – ω² = (–ω²) – ω² = –2ω².

Step 3: Now raise to the third power:

\( (-2\omega^2)^3 = (-2)^3 (\omega^2)^3 = -8\,\omega^6 \)

But ω⁶ = (ω³)² = 1² = 1, so

  (1 + ω – ω²)³ = –8.

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Example 6:
Evaluate

  (1 + ω + ω²) + (1 + ω – ω²) + (1 – ω + ω²)

Since 1 + ω + ω² = 0, the sum becomes:

  0 + (1 + ω – ω²) + (1 – ω + ω²)
  = 1 + ω – ω² + 1 – ω + ω²
  = 2

Thus, the value is 2.

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Roots and Coefficients of a Quadratic Equation

The Relationship

The roots of a quadratic equation and their coefficients have following relation.

If α and β are the roots of the quadratic equation

  ax² + bx + c = 0,

then it can be factored as:

  a(x – α)(x – β) = 0.

Expanding, we get:

  ax² – a(α + β)x + aαβ = 0.

By comparing coefficients, we deduce:

• Sum of the roots:
    α + β = –b/a
• Product of the roots:
    αβ = c/a

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Finding the Sum and Product Without Solving

Example 7:
For 3x² + 5x – 7 = 0, we have:
  a = 3, b = 5, c = –7

• Sum: α + β = –5/3
• Product: αβ = –7/3

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Example 8:
For x² – 9 = 0, rewrite it as:

  x² + 0x – 9 = 0
  (a = 1, b = 0, c = –9)

• Sum: α + β = 0
• Product: αβ = –9

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Finding Unknowns Using Conditions on the Roots

By applying some conditions on the roots of a quadratic equation we can find unknown values.

(a) Sum Equals a Multiple of the Product

Example 9:
For

  x² – (k + 6)x + 2(2k – 1) = 0

the condition is that the sum of the roots equals half of their product.

• Here, a = 1, b = –(k + 6), c = 2(2k – 1)
• Sum: α + β = k + 6
• Product: αβ = 2(2k – 1)

Set up the condition:

\( k + 6 = \frac{1}{2} \times 2(2k – 1) \)

Simplify:

  k + 6 = 2k – 1 ⟹ k = 7

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(b) Sum of the Squares of the Roots

Example 10:
For x² + 2x + k = 0, if

  α² + β² = 40

we have:

• Sum: α + β = –2
• Product: αβ = k

Recall that

\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta \)

Substitute:

\( 40 = (-2)^2 – 2k \quad \Rightarrow \quad 40 = 4 – 2k \)

Solve:

  –2k = 36 ⟹ k = –18

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(c) Roots Differ by a Given Number

Example 11:
For x² – 4x + k = 0, if the roots differ by 3 (i.e. |α – β| = 3):

• Sum: α + β = 4

Assume α – β = 3. Then:

  (α + β) + (α – β) = 4 + 3 = 7 ⟹ 2α = 7 ⟹ α = 7/2
  β = 4 – (7/2) = 1/2

The product is:

  αβ = (7/2) × (1/2) = 7/4, so k = 7/4.

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(d) Roots Satisfy a Given Relation

Example 12:
For x² – 5x + k = 0, if the roots satisfy 3α + 2β = 16:

• Sum: α + β = 5

Express β = 5 – α, then substitute into the relation:

3α + 2(5 − α) = 16

Simplify:

  3α + 10 – 2α = 16 ⟹ α = 6
  Then, β = 5 – 6 = –1
  So, αβ = 6 × (–1) = –6 ⟹ k = –6.

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(e) Both Sum and Product Are Given a Specific Value

Example 13:
For

  x² – kx + (2k – 3) = 0

if both the sum and product of the roots are required to be 2, then:

• Sum: α + β = k, so k must equal 2.
• Product: αβ = 2k – 3. With k = 2, this gives 2(2) – 3 = 1, which is not 2.

Since the two conditions lead to different values of k, it is not possible for both the sum and product of the roots to equal 2 for any k.

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Symmetric Functions of the Roots

A function involving the roots of a quadratic equation is considered symmetric if its value remains unchanged when you interchange the roots. For example:

• α + β is symmetric (since α + β = β + α).
• αβ is symmetric.
• α² + β² is symmetric.
• However, α – β is not symmetric because swapping the roots gives –(α – β).

Any expression that can be rewritten in terms of (α + β) and (αβ) is symmetric.

Examples:

• α² + β²:
     (α + β)² – 2αβ
  α² + β² = (α + β)² − 2αβ
    
• α³ + β³:
    α³ + β³ = (α + β)³ − 3αβ(α + β)
• (α – β)²:
    (α − β)² = (α + β)² − 4αβ
• 1/α + 1/β:
    (α + β)/αβ
    

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Forming a Quadratic Equation

The roots of a quadratic equation are essential; they allow us to construct the equation itself. These roots represent the solutions, the values that make the equation true; they are the answers we seek. Furthermore, we have the ability to transform or modify these roots in various ways. By altering the roots, we can generate entirely new quadratic equations with different characteristics.

1. From Given Roots

If you know the two roots, α and β, the quadratic equation having those roots is:

  (x – α)(x – β) = 0

Expanding:

  x² – (α + β)x + αβ = 0

So, the formula is:

  x² – (sum of the roots)x + (product of the roots) = 0

Example 14:
Form a quadratic equation with roots 2 and 3.

• Sum = 2 + 3 = 5
• Product = 2 × 3 = 6
• Equation: x² – 5x + 6 = 0

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Example 15:
Form a quadratic equation with roots –2 and 4.

• Sum = –2 + 4 = 2
• Product = –2 × 4 = –8
• Equation: x² – 2x – 8 = 0

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2. Forming Equations with Transformed Roots

Sometimes, you may want to form a new quadratic equation when the roots are transformed from the original roots α and β.

(a) New Roots: 2α + 1 and 2β + 1

• New Sum: 2α + 1 + 2β + 1 = 2(α + β) + 2
• New Product: (2α + 1)(2β + 1) = 4αβ + 2(α + β) + 1

If the original equation is ax² + bx + c = 0, then
  α + β = –b/a and αβ = c/a.

Thus, the new quadratic (after clearing fractions) becomes:

ax² + (2b − 2a)x + (4c − 2b + a) = 0

Example 16:
For x² – 5x + 6 = 0, we have α + β = 5 and αβ = 6.

• New Sum = 2(5) + 2 = 12
• New Product = 4(6) + 2(5) + 1 = 24 + 10 + 1 = 35
• New equation: x² – 12x + 35 = 0

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(b) New Roots: α² and β²

• New Sum: α² + β² = (α + β)² – 2αβ
• New Product: α²β² = (αβ)²

Expressed in terms of a, b, and c:

• New sum: (b² – 2ac)/a²
• New product: c²/a²

Multiplying by a², the new equation is:

a²x² − (b² − 2ac)x + c² = 0

Example 17:
For x² – 5x + 6 = 0 (with a = 1, b = –5, c = 6):

• New Sum = 5² – 2×6 = 25 – 12 = 13
• New Product = 6² = 36
• New equation: x² – 13x + 36 = 0

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(c) New Roots: 1/α and 1/β

• New Sum: 1/α + 1/β = (α + β)/αβ
• New Product: 1/(αβ)

Since α + β = –b/a and αβ = c/a:

• New Sum = –b/c
• New Product = a/c

The new quadratic (after multiplying by c) is:

cx² + bx + a = 0

Example 18:
For x² – 5x + 6 = 0:

• New Sum = 5/6 and New Product = 1/6
• New equation (after clearing fractions): 6x² – 5x + 1 = 0

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(d) New Roots: α/β and β/α

• New Sum: α/β + β/α = (α² + β²)/αβ = [(α + β)² – 2αβ]/αβ
• New Product: (α/β)(β/α) = 1

In terms of a, b, and c, the new sum becomes:

  (b² – 2ac)/(ac)

Thus, the new quadratic is:
acx² − (b² − 2ac)x + ac = 0

Example 19:
For x² – 5x + 6 = 0 (a = 1, b = –5, c = 6):

• New Sum = (25 – 12) / 6 = 13/6
• New Product = 1
• New equation (multiplying by 6): 6x² – 13x + 6 = 0

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(e) New Roots: (α + β) and (1/α + 1/β)

• One new root is α + β.
• The other is 1/α + 1/β = (α + β)/αβ.

Let R = α + β and S = (α + β)/αβ.

Then:

• New Sum: R + S = (α + β) [1 + 1/(αβ)]
• New Product: R × S = (α + β)²/αβ

Substituting α + β = –b/a and αβ = c/a, we get:

• New Sum = –(b/a)[1 + (a/c)] = –b(c + a)/(ac)
• New Product = b²/(ac)

Thus, the new quadratic (after multiplying by ac) is:
acx² + b(a + c)x + b² = 0

Example 20:
For x² – 5x + 6 = 0:

• α + β = 5 and αβ = 6
• One new root is 5, the other is 5/6.
• Their sum is 5 + 5/6 = 35/6 and their product is 25/6.
• The new equation (after clearing fractions) is: 6x² – 35x + 25 = 0

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Final Thoughts

This guide on the roots of a quadratic equation has walked you through everything from the basic form of a quadratic equation to using the discriminant to determine the nature of its roots. We also explored the fascinating cube roots of unity, explained the relationships between roots and coefficients, and demonstrated how to form new quadratic equations when the roots are transformed. You can find fully solved exercises on these topics here.

For a deeper understanding of how roots of a quadratic equation are represented graphically, check out this detailed article on Cuemath.