Roots of Quadratic Equations

Master the concept of the roots of quadratic equations easily.

📚 Further Maths 🎯 Difficulty: ⭐ ⏱️ Reading time: 15 minutes 📋 Edexcel

📚 Introduction

Roots of a quadratic equation are the values of $x$ that make the equation equal to zero. If a quadratic has roots $\alpha$ and $\beta$, then it can be written in factorized form as $a(x-\alpha)(x-\beta) = 0$ where $a$ is a constant.
When we expand the brackets correctly: $$a(x-\alpha)(x-\beta) = a\left(x^2 - (\alpha + \beta)x + \alpha\beta\right) = ax^2 - a(\alpha + \beta)x + a\alpha\beta$$
In this topic, we take a look at the relationship between the roots $\alpha$ and $\beta$ and the coefficients of the standard form $ax^2 + bx + c = 0$.

$\alpha, \beta$ are just the greek letters 'alpha' and 'beta'. They are commonly used to represent roots of equations.

🎯 Learning Objectives

Find the roots of quadratic equations using various methods
Understand the relationship between the coefficients and roots
Remember the Vietas formulae

Remember this:
The roots of a quadratic can either be:

Two, real and different roots
Two, real and the same but repeating roots
Two, complex roots (non-real)

This is determined by the discriminant $b^2 - 4ac$, which you should already know.

🔑 Key Concepts

Formulae for roots

The formula for the roots of a quadratic equation is given by:

\text{Sum of roots: }\alpha + \beta = -\frac{b}{a}$$ $$\text{Product of roots: }\alpha \beta = \frac{c}{a}$$ $$\alpha, \beta = \text{The roots of the quadratic equation}$$ $$a, b, c = \text{The coefficients of the quadratic equation}

These relationships are derived from the standard form of a quadratic equation $ax^2 + bx + c = 0$. The entire derivation:
Starting with the factorized form: $$a(x - \alpha)(x - \beta) = 0$$ Expanding this gives: $$ax^2 - a(\alpha + \beta)x + a\alpha\beta = 0$$ Comparing coefficients with $ax^2 + bx + c = 0$ leads to: $$b = -a(\alpha + \beta) \quad \Rightarrow \quad \boxed{\alpha + \beta =-\frac{b}{a}}$$ $$c = a\alpha\beta \quad \Rightarrow \quad \boxed{\alpha\beta = \frac{c}{a}}$$

📝 Worked Examples

📋 Example 1: Simple example:

Question: a quadratic is given: $$15z^2 + 5z - 2 = 0$$ The roots of this quadratic are $\alpha$ and $\beta$. find $\alpha \beta$ and $\alpha + \beta$.

Solution:

Step 1:

Recall the formulae:

\alpha + \beta = -\frac{b}{a}$$ $$\alpha \beta = \frac{c}{a}

Step 2:

Now just plug the values.

\text{We have } a = 15, b = 5, c = -2$$ $$\text{So: } \alpha + \beta = -\frac{5}{15} = -\frac{1}{3}$$ $$\text{And: } \alpha \beta = \frac{-2}{15} = -\frac{2}{15}

💡 Tip:

The quadratic can be written differently. $$\text{If } ax^2 + bx + c = a(x - \alpha)(x - \beta)$$ $$\text{Then dividing by } a: \quad x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$

📋 Example 2: Problem involving roots:

The roots of $$f(x) = ax^2 + bx + c$$ are $\alpha$ and $\beta$. Find $a$, $b$, $c$ when: $$\alpha = - \frac{3}{2}, \beta = \frac{5}{4}$$

Solution:

Step 1:

Recall the formulae:

\alpha + \beta = -\frac{b}{a}$$ $$\alpha \beta = \frac{c}{a}$$ $$\text{And in this case: }$$ $$\alpha = - \frac{3}{2}, \quad \beta = \frac{5}{4}

Step 2:

Lets solve $\alpha + \beta$:

\alpha + \beta = -\frac{3}{2} + \frac{5}{4} = -\frac{6}{4} + \frac{5}{4} = -\frac{1}{4}$$ $$\text{Since } \alpha+\beta = -\frac{b}{a}, \text{ we have: } -\frac{b}{a} = -\frac{1}{4} \Rightarrow b = \frac{a}{4}

Step 3:

Lets now solve $\alpha \beta$:

\alpha \beta = -\frac{3}{2} \times \frac{5}{4} = -\frac{15}{8}

Step 4:

Now, since $\alpha \beta = \frac{c}{a}$, and we have $\alpha \beta = -\frac{15}{8}$, we can write:

x^2 + \frac{1}{4}x - \frac{15}{8}$$ $$\text{Multiplying through by 8 to clear the fractions: }$$ $$8x^2 + 2x - 15 \text{ Which is in the form } ax^2 + bx + c$$ $$\boxed{\text{So: } a = 8, b = 2, c = -15}

⚠️ Common Mistakes

🚫 Mistake 1: Forgetting the formulae

What students often do wrong: Just remember the formulae for sum and product of roots. They are easy to forget. $$\alpha + \beta = -\frac{b}{a}$$ $$\alpha \beta = \frac{c}{a}$$

💪 Practice Problems

Try these problems to test your understanding:

🎯 Practice Questions A

Part 1: α and β are the roots of the quadratic equation $7x^2 - 3x + 1 = 0$. Without solving the equation, find the values of:
A. $\alpha + \beta$
B. $\alpha\beta$
C. $\frac{1}{\alpha} + \frac{1}{\beta}$
D. $\alpha^2 + \beta^2$

Part 2: α and β are the roots of the quadratic equation $6x^2 - 9x + 2 = 0$. Without solving the equation, find the values of:
A. $\alpha + \beta$
B. $\alpha^2 \times \beta^2$
C. $\frac{1}{\alpha} + \frac{1}{\beta}$
D. $\alpha^3 + \beta^3$

🔍 Show Solution

Part 1: For $7x^2 - 3x + 1 = 0$, we have $a = 7$, $b = -3$, $c = 1$ $$\text{A. } \alpha + \beta = -\frac{b}{a} = -\frac{(-3)}{7} = \frac{3}{7}$$ $$\text{B. } \alpha\beta = \frac{c}{a} = \frac{1}{7}$$ $$\text{C. } \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{\frac{3}{7}}{\frac{1}{7}} = 3$$ $$\text{D. } \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{3}{7}\right)^2 - 2 \cdot \frac{1}{7} = \frac{9}{49} - \frac{2}{7} = \frac{9 - 14}{49} $$ $$= -\frac{5}{49}$$

Part 2: For $6x^2 - 9x + 2 = 0$, we have $a = 6$, $b = -9$, $c = 2$ $$\text{A. } \alpha + \beta = -\frac{b}{a} = -\frac{(-9)}{6} = \frac{9}{6} = \frac{3}{2}$$ $$\text{B. } \alpha^2 \times \beta^2 = (\alpha\beta)^2 = \left(\frac{c}{a}\right)^2 = \left(\frac{2}{6}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}$$ $$\text{C. } \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{\frac{3}{2}}{\frac{1}{3}} = \frac{3}{2} \times 3 = \frac{9}{2}$$ $$\text{D. } \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = \left(\frac{3}{2}\right)^3 - 3 \cdot \frac{1}{3} \cdot \frac{3}{2} = \frac{27}{8} - \frac{3}{2} $$ $$= \frac{27 - 12}{8} = \frac{15}{8}$$

🎯 Practice Questions B

Question 1: The roots of the quadratic equation $px^2 + qx + r = 0$ are $\alpha = \frac{2 + \sqrt{5}}{3}$ and $\beta = \frac{2 - \sqrt{5}}{3}$. Find integer values of $p$, $q$ and $r$.

Question 2: One root of the quadratic $ax^2 + bx + c = 0$ is $\alpha = 3 - 2i$.
a) Write down the other root $\beta$.
b) If $a = 2$, find the integers $b$ and $c$.

Question 3: Given that $kx^2 + (k+4)x - 6 = 0$, find the integer value of $k$ if the product of the roots is $-2$.

🔍 Show Solution

Question 1 Solution: $$\alpha + \beta = \frac{2 + \sqrt{5}}{3} + \frac{2 - \sqrt{5}}{3} = \frac{4}{3}$$ $$\alpha \beta = \frac{2 + \sqrt{5}}{3} \times \frac{2 - \sqrt{5}}{3} = \frac{(2 + \sqrt{5})(2 - \sqrt{5})}{9} = \frac{4 - 5}{9} = -\frac{1}{9}$$ $$\text{Using } \alpha + \beta = -\frac{q}{p} \text{ and } \alpha\beta = \frac{r}{p}:$$ $$-\frac{q}{p} = \frac{4}{3} \Rightarrow q = -\frac{4p}{3}$$ $$\frac{r}{p} = -\frac{1}{9} \Rightarrow r = -\frac{p}{9}$$ $$\text{For integer values, let } p = 9: \text{ then } q = -12, r = -1$$ $$\boxed{p = 9, q = -12, r = -1}$$

Question 2 Solution: $$\text{a) Since complex roots come in conjugate pairs: } \boxed{\beta = 3 + 2i}$$ $$\text{b) } \alpha + \beta = (3 - 2i) + (3 + 2i) = 6$$ $$\alpha \beta = (3 - 2i)(3 + 2i) = 9 - (2i)^2 = 9 + 4 = 13$$ $$\text{With } a = 2: \alpha + \beta = -\frac{b}{a} = -\frac{b}{2} = 6 \Rightarrow b = -12$$ $$\alpha \beta = \frac{c}{a} = \frac{c}{2} = 13 \Rightarrow c = 26$$ $$\boxed{b = -12, c = 26}$$

Question 3 Solution: $$\text{For } kx^2 + (k+4)x - 6 = 0: a = k, b = k+4, c = -6$$ $$\text{Product of roots } = \frac{c}{a} = \frac{-6}{k} = -2$$ $$\frac{-6}{k} = -2 \Rightarrow k = 3$$ $$\boxed{k = 3}$$

📋 Summary

🎯 Key Takeaways

The sum of roots formula: $\alpha + \beta = -\frac{b}{a}$
The product of roots formula: $\alpha \beta = \frac{c}{a}$
These formulae work for any quadratic equation in the form $ax^2 + bx + c = 0$
You can find coefficients when given the roots using these relationships
Complex roots come in conjugate pairs for real coefficient quadratics
These relationships are part of Vieta's formulae for polynomials

📚 What's Next?

Now that you've mastered quadratic roots, it's time to explore the more complex world of cubic equations. Learn about the sum and product relationships for three roots and tackle more challenging polynomial problems.

🚀 Continue to Roots of Cubic Equations

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