Chapter 6 — Quadratic Equations
Topics in this chapter:
6.1 What’s a Quadratic?
A quadratic equation is a polynomial equation of degree 2:
\[ax^2 + bx + c = 0 \qquad (a \neq 0)\]where $a$, $b$, $c$ are constants. Quadratics arise everywhere — from physics (projectile motion) to optimization.
The graph of $y = ax^2 + bx + c$ is a parabola.
6.2 Factoring Quadratics
The Distributive Property and FOIL
To solve $ax^2 + bx + c = 0$ by factoring, we want to write the left side as a product of two linear terms.
Recall: $(x + r)(x + s) = x^2 + (r+s)x + rs$.
So to factor $x^2 + bx + c$, find two numbers $r$ and $s$ such that $r + s = b$ and $rs = c$.
Example 6-1: Factor $x^2 + 4x + 3$.
Solution: We need $r + s = 4$ and $rs = 3$. The pair $(r,s) = (1, 3)$ works.
\[x^2 + 4x + 3 = (x+1)(x+3)\]So the equation $x^2 + 4x + 3 = 0$ has roots $x = -1$ and $x = -3$.
Three Special Factorizations
| Pattern | Factorization | Example |
|---|---|---|
| Zero root | $ax^2 + bx = x(ax + b)$ | $3x^2 + 6x = 3x(x+2) = 0 \implies x = 0$ or $x = -2$ |
| Difference of squares | $x^2 - a^2 = (x-a)(x+a)$ | $x^2 - 9 = (x-3)(x+3) = 0 \implies x = \pm 3$ |
| Perfect square | $x^2 \pm 2ax + a^2 = (x \pm a)^2$ | $x^2 + 6x + 9 = (x+3)^2 = 0 \implies x = -3$ (double root) |
Example 6-2: Factor $2x^2 - 5x - 3$.
Solution: We need $r \cdot s = (2)(-3) = -6$ and $r + s = -5$. The pair is $(-6, 1)$.
Split: $2x^2 - 6x + x - 3 = 2x(x-3) + 1(x-3) = (2x+1)(x-3)$.
Roots: $x = -\frac{1}{2}$ and $x = 3$.
Exercise 6-1: Factor completely and find the roots:
i. $x^2 - 5x + 6 = 0$ ii. $2y^2 + 5y - 3 = 0$ iii. $3z^2 + 7z + 2 = 0$
Example 6-3: Solve $56x^2 + 19x - 10 = 0$.
Solution: We need two numbers that multiply to $56(-10) = -560$ and sum to $19$.
Since $560 = 4 \cdot 140 = 4 \cdot 4 \cdot 35 = 16 \cdot 35$, and $35 - 16 = 19$, we use $(35, -16)$.
\[56x^2 + 35x - 16x - 10 = 7x(8x+5) - 2(8x+5) = (7x-2)(8x+5) = 0\]Roots: $x = \frac{2}{7}$ or $x = -\frac{5}{8}$.
Example 6-4: Solve $x^4 - 1 = 0$.
Solution: Difference of squares:
\[x^4 - 1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1) = 0\]Real roots: $x = \pm 1$. (Over the complex numbers: also $x = \pm i$.)
Exercise 6-2: Solve by factoring:
i. $3x^2 + 12x = 0$ ii. $4x^2 - 16 = 0$ iii. $x^2 + 10x + 25 = 0$
Exercise 6-3: Factor and solve $x^4 - 16 = 0$ over the reals.
Exercise 6-4: Solve without the quadratic formula: $x^2 + 2bx = c$. (Hint: complete the square.)
6.3 The Quadratic Formula
Deriving the Formula by Completing the Square
Starting from $ax^2 + bx + c = 0$ (with $a \neq 0$):
\[x^2 + \frac{b}{a}x = -\frac{c}{a}\]Add $\left(\frac{b}{2a}\right)^2$ to both sides:
\[\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\]Taking the square root:
The Quadratic Formula:
\[\boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\]The Discriminant
The discriminant $\Delta = b^2 - 4ac$ determines the nature of the roots:
| Discriminant | Nature of Roots |
|---|---|
| $\Delta > 0$ | Two distinct real roots |
| $\Delta = 0$ | One repeated (double) real root |
| $\Delta < 0$ | Two complex conjugate roots (no real roots) |
Example 6-5: Solve $3x^2 + 7x + 2 = 0$.
Solution:
\[x = \frac{-7 \pm \sqrt{49 - 24}}{6} = \frac{-7 \pm \sqrt{25}}{6} = \frac{-7 \pm 5}{6}\]$x = -\frac{1}{3}$ or $x = -2$.
Check by factoring: $(3x+1)(x+2) = 0$ gives the same roots. ✓
Vieta’s Formulas
If $r$ and $s$ are roots of $ax^2 + bx + c = 0$, then:
\[r + s = -\frac{b}{a}, \qquad r \cdot s = \frac{c}{a}\]These are called Vieta’s formulas. They let you compute symmetric functions of the roots without solving.
Application: Knowing the sum and product of roots, you can reconstruct the quadratic:
\[x^2 - (r+s)x + rs = 0\]Example 6-6: If $r$ and $s$ are roots of $x^2 - 5x + 3 = 0$, find $r^2 + s^2$.
Solution: By Vieta’s: $r + s = 5$ and $rs = 3$.
\[r^2 + s^2 = (r+s)^2 - 2rs = 25 - 6 = 19\]Example 6-7: Find a quadratic with integer coefficients whose roots are $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$.
Solution: Sum $= 3$, product $= \frac{9 - 5}{4} = 1$.
\[x^2 - 3x + 1 = 0\]Exercise 6-5: Use the quadratic formula to solve:
i. $x^2 - 7x + 10 = 0$ ii. $3x^2 - 4x - 2 = 0$ iii. $x^2 + x + 1 = 0$
Exercise 6-6: For each equation in Exercise 6-5, state the discriminant and describe the nature of the roots.
Exercise 6-7: If $r$ and $s$ are roots of $2x^2 + 7x - 3 = 0$, find $\frac{1}{r} + \frac{1}{s}$ and $r^2 + s^2$.
Exercise 6-8: Write a quadratic with roots $3 + 2i$ and $3 - 2i$.
6.4 Variations on a Theme
6.4.1 Rearrangements
Some equations don’t look quadratic at first, but become quadratic after rearrangement.
Example 6-8: Solve $\;\dfrac{3}{x-1} + \dfrac{4}{x+3} = 1$.
Solution: Common denominator $(x-1)(x+3)$:
\[3(x+3) + 4(x-1) = (x-1)(x+3)\] \[3x + 9 + 4x - 4 = x^2 + 2x - 3\] \[7x + 5 = x^2 + 2x - 3 \implies x^2 - 5x - 8 = 0\] \[x = \frac{5 \pm \sqrt{25 + 32}}{2} = \frac{5 \pm \sqrt{57}}{2}\]Warning — Extraneous Roots: When multiplying both sides by an expression containing the variable, you may introduce false solutions. Always check your answers in the original equation!
Example 6-9: Solve $\;\sqrt{x+7} = x - 5$.
Solution: Square both sides: $x + 7 = x^2 - 10x + 25$, so $x^2 - 11x + 18 = 0$, giving $x = 2$ or $x = 9$.
Check: $x = 2$: $\sqrt{9} = 3 \neq 2 - 5 = -3$. Extraneous!
$x = 9$: $\sqrt{16} = 4 = 9 - 5$. ✓
Answer: $x = 9$ only.
Example 6-10: Solve $\;\sqrt{x+5} - \sqrt{x} = 1$.
Solution: Isolate one radical: $\sqrt{x+5} = 1 + \sqrt{x}$.
Square: $x + 5 = 1 + 2\sqrt{x} + x$, so $4 = 2\sqrt{x}$, giving $\sqrt{x} = 2$, $x = 4$.
Check: $\sqrt{9} - \sqrt{4} = 3 - 2 = 1$. ✓
6.4.2 Substitutions
Some equations become quadratic with the right substitution.
Example 6-11: Solve $\;2^{2x} - 3 \cdot 2^x + 2 = 0$.
Solution: Let $u = 2^x$, so $u^2 - 3u + 2 = 0$, giving $(u-1)(u-2) = 0$.
$u = 1 \implies 2^x = 1 \implies x = 0$.
$u = 2 \implies 2^x = 2 \implies x = 1$.
Example 6-12: Solve $\;x^4 - 3x^2 + 2 = 0$.
Solution: Let $u = x^2$: $u^2 - 3u + 2 = 0$, so $(u-1)(u-2) = 0$.
$u = 1 \implies x = \pm 1$. $\;u = 2 \implies x = \pm\sqrt{2}$.
Example 6-13: Solve $\;\left(x - \dfrac{1}{x}\right) - 4\left(x - \dfrac{1}{x}\right)^{1/2} + 4 = 0$.
Solution: Let $u = \left(x - \frac{1}{x}\right)^{1/2}$, so $u^2 - 4u + 4 = 0$ giving $(u-2)^2 = 0$, $u = 2$.
Then $x - \frac{1}{x} = 4$, so $x^2 - 4x - 1 = 0$:
\[x = \frac{4 \pm \sqrt{16 + 4}}{2} = 2 \pm \sqrt{5}\]Check: $u = \sqrt{x - 1/x}$ requires $x - 1/x \geq 0$. For $x = 2 + \sqrt{5} > 0$: works. For $x = 2 - \sqrt{5} < 0$: $x - 1/x = 4$ still, and $\sqrt{4} = 2$. ✓ Both are valid.
Exercise 6-9: Solve:
i. $\sqrt{2x+3} = x$ ii. $4^x - 6 \cdot 2^x + 8 = 0$ iii. $x^6 - 9x^3 + 8 = 0$
Exercise 6-10: Find all $x$ such that $\frac{1}{x} - \frac{1}{x+1} = \frac{1}{4}$.
6.5 Square Roots of Irrationals and Imaginaries
Can we simplify $\sqrt{3 + 2\sqrt{2}}$?
Method: Assume $\sqrt{a + b\sqrt{c}} = \sqrt{p} + \sqrt{q}$.
Square: $a + b\sqrt{c} = p + q + 2\sqrt{pq}$.
Match rational and irrational parts:
\[p + q = a, \qquad 2\sqrt{pq} = b\sqrt{c} \implies pq = \frac{b^2 c}{4}\]Solve this system for $p$ and $q$.
Example 6-14: Simplify $\sqrt{3 + 2\sqrt{2}}$.
Solution: $p + q = 3$ and $pq = 2$. These are roots of $t^2 - 3t + 2 = 0$, giving $t = 1, 2$.
\[\sqrt{3 + 2\sqrt{2}} = \sqrt{2} + 1\]Check: $(\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}$. ✓
Example 6-15: Simplify $\sqrt{5 + 12i}$ where $i = \sqrt{-1}$.
Solution: Assume $\sqrt{5 + 12i} = a + bi$ (with real $a, b$).
$(a + bi)^2 = a^2 - b^2 + 2abi = 5 + 12i$.
So $a^2 - b^2 = 5$ and $2ab = 12$, i.e. $ab = 6$.
From $ab = 6$: $b = 6/a$. Then $a^2 - 36/a^2 = 5$, giving $a^4 - 5a^2 - 36 = 0$.
$(a^2 - 9)(a^2 + 4) = 0 \implies a^2 = 9 \implies a = 3$ (taking positive root).
$b = 6/3 = 2$.
\[\sqrt{5 + 12i} = 3 + 2i\]Check: $(3+2i)^2 = 9 + 12i - 4 = 5 + 12i$. ✓
Exercise 6-11: Simplify $\sqrt{6 - 4\sqrt{2}}$ and $\sqrt{-3 + 4i}$.
6.6 Beyond Quadratics
Polynomials of higher degree:
- Cubic ($n=3$): $ax^3 + bx^2 + cx + d = 0$
- Quartic ($n=4$): $ax^4 + bx^3 + cx^2 + dx + e = 0$
- A degree-$n$ polynomial has at most $n$ roots.
Vieta’s for cubics: For $x^3 + px^2 + qx + r = 0$ with roots $\alpha, \beta, \gamma$:
\[\alpha + \beta + \gamma = -p, \qquad \alpha\beta + \alpha\gamma + \beta\gamma = q, \qquad \alpha\beta\gamma = -r\]The Big Picture — Cubics and Beyond
The general cubic equation was first solved by Tartaglia (c. 1535) using a clever substitution. The formula was published by Cardano (1545). The quartic was solved by Ferrari, Cardano’s student.
Évariste Galois (1811–1832) proved that no general algebraic formula exists for degree $\geq 5$, founding Galois theory — one of the great achievements of mathematics.
Exercises & Problems
End-of-Chapter Problems
Problem 110. Solve $x^2 - 3x + 2 = 0$.
Problem 111. Find the sum and product of the roots of $2x^2 + 3x - 5 = 0$. (Use Vieta’s.)
Problem 112. If $x^2 - 5x + c = 0$ has a double root, find $c$.
Problem 113. Solve $x^4 - 5x^2 + 4 = 0$ in the reals.
Problem 114. If $r$ and $s$ are roots of $x^2 + 5x + 3 = 0$, find $r^2 + s^2$ and $\frac{1}{r} + \frac{1}{s}$. (AHSME 1955)
Problem 115. Solve $\frac{1}{x^2-10x-29}+\frac{1}{x^2-10x-45}-\frac{2}{x^2-10x-69}=0$. (AHSME 1983)
Problem 116. If $2^{2x} - 2^{x+1} - 32 = 0$, find $x$.
Problem 117. Find all values of $x$ such that $\sqrt{3x+1} = 1 + \sqrt{x+4}$.
Problem 118. Simplify $\sqrt{7 + 4\sqrt{3}}$.
Problem 119. If $r$ and $s$ are roots of $x^2 = 5x - 2$, compute $r^3 + s^3$.
Problem 120. Solve $(x + 1)(x + 2)(x + 3)(x + 4) = 120$. (Hint: pair the factors.)
| Problem 121. How many real solutions does $ | x^2 + 2x - 1 | = | x - 3 | $ have? |
Problem 122. For what values of $k$ does $x^2 + kx + k + 2 = 0$ have two real solutions? (AHSME 1961)
Problem 123. If the sum of the roots of $ax^2 + bx + c = 0$ equals the sum of the squares of the roots, express $b$ in terms of $a$ and $c$.
Problem 124. Solve $\sqrt{2x + 1} + \sqrt{x - 3} = 2\sqrt{x}$ for $x$.
Problem 125. If $r, s$ are roots of $ax^2 + bx + c = 0$, find $r^2s + rs^2$ in terms of $a, b, c$.
| Problem 126. Find all $x$ satisfying $ | x^2 - 4 | < 5$. |
Problem 127. If one root of $x^2 + px + q = 0$ is three times the other, express $p^2$ in terms of $q$. (MAΘ 1991)
Problem 128. Find an equation whose roots are the squares of the roots of $x^2 + x - 3 = 0$. (AHSME 1954)
Problem 129. In the equation $x^2 + ax + b = 0$, one root is $\sqrt{3} + \sqrt{2}$. Find a rational value of $b$ that makes this possible. (AHSME 1952)
Problem 130. Write $\sqrt[4]{97 - 56\sqrt{3}}$ in the form $a + b\sqrt{3}$.
Problem 131. The equation $x^2 + bx + c = 0$ has two roots that sum to 6 and whose absolute values sum to 10. Find $b$ and $c$.
Problem 132. Find all real $x$ such that $x^4 + 4x^3 + 6x^2 + 4x - 3 = 0$.
Problem 133. If $a + b + c = 0$, show that $a^2 + b^2 + c^2 = -2(ab + bc + ca)$ and $(a^2 + b^2 + c^2)^2 = 2(a^4 + b^4 + c^4)$. (Various sources)
Problem 134. Determine, without solving, whether all roots of $x^2 + 2ax + 2a^2 + 4a - 3 = 0$ are real.
Problem 135. If $r_1$ and $r_2$ are roots of $x^2 + px + q = 0$, then $r_1^2 + r_2^2$ equals what expression? (AHSME 1951)
Problem 136. Solve $(x - a)(x - b) = (r - a)(r - b)$ for $x$ given $r \neq x$.
Problem 137. The absolute value of the difference of the roots of $x^2 + px + 1 = 0$. Express in terms of $p$, given $p^2 > 4$.