Quadratic equations with real coefficients

In Class X, the equation $x^2 + 1 = 0$ had "no solution". With complex numbers in hand, every quadratic with real coefficients has exactly two complex roots (counting a repeated root twice). This subtopic puts the quadratic formula in its full Class-XI form.

The quadratic formula

For $ax^2 + bx + c = 0$ with real $a \ne 0, b, c$: $\boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.}$

The expression $b^2 - 4ac$ is called the discriminant, denoted $\Delta$ or $D$. Its sign decides the nature of the roots:

• $\Delta > 0$: two distinct real roots.
• $\Delta = 0$: one repeated real root (a double root).
• $\Delta < 0$: two complex conjugate roots: $x = \dfrac{-b \pm i\sqrt{|\Delta|}}{2a}$.

In all three cases, we now write the formula uniformly, allowing complex square roots.

Derivation by completing the square

Divide by $a$: $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$. Complete the square: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}.$ Take square roots (allowed in $\mathbb{C}$): $x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a},$ so $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \qed$

Conjugate root theorem

If a quadratic with real coefficients has a complex root $\alpha + i\beta$ with $\beta \ne 0$, then its conjugate $\alpha - i\beta$ is also a root. Reason: taking conjugates of $a x^2 + bx + c = 0$ and using that $a, b, c$ are real gives $a \bar{x}^2 + b \bar{x} + c = 0$, so $\bar{x}$ is also a root.

This is why complex roots of a real quadratic always come in conjugate pairs.

Sum and product of roots (Vieta's formulas)

If $\alpha, \beta$ are the two roots of $ax^2 + bx + c = 0$, then $\alpha + \beta = -\frac{b}{a}, \qquad \alpha \beta = \frac{c}{a}.$

This works whether the roots are real or complex. Direct proof: factor $a x^2 + bx + c = a(x - \alpha)(x - \beta) = ax^2 - a(\alpha + \beta) x + a\alpha\beta$, then match coefficients.

Constructing a quadratic from roots

Given roots $\alpha, \beta$: $x^2 - (\alpha + \beta) x + \alpha\beta = 0$ is a quadratic with those roots. Any scalar multiple is equivalent.

Worked examples

Example 1. Solve $x^2 - 5x + 6 = 0$.

$\Delta = 25 - 24 = 1$. $x = \dfrac{5 \pm 1}{2} = 3, 2$.

Example 2. Solve $x^2 + 2x + 5 = 0$.

$\Delta = 4 - 20 = -16$. $x = \dfrac{-2 \pm \sqrt{-16}}{2} = \dfrac{-2 \pm 4i}{2} = -1 \pm 2i$.

Example 3. Find the quadratic equation whose roots are $3 + i$ and $3 - i$.

Sum $= 6$, product $= 9 + 1 = 10$. Equation: $x^2 - 6x + 10 = 0$.

Example 4. For what values of $k$ does $x^2 - 6x + k = 0$ have equal roots?

Equal roots iff $\Delta = 36 - 4k = 0 \Rightarrow k = 9$.

Example 5 (harder). If $\alpha$ and $\beta$ are the roots of $x^2 - 4x + 5 = 0$, find a quadratic with roots $\alpha^2 + 1, \beta^2 + 1$.

$\alpha + \beta = 4$, $\alpha\beta = 5$. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 16 - 10 = 6$.

Sum of new roots: $\alpha^2 + \beta^2 + 2 = 8$. Product of new roots: $(\alpha^2 + 1)(\beta^2 + 1) = \alpha^2 \beta^2 + \alpha^2 + \beta^2 + 1 = 25 + 6 + 1 = 32$.

New quadratic: $x^2 - 8x + 32 = 0$.

Try it yourself

1. Solve $x^2 - 7x + 12 = 0$.
2. Solve $x^2 + 4x + 13 = 0$.
3. Solve $2x^2 + x + 3 = 0$.
4. Find $k$ such that $x^2 + kx + 9 = 0$ has equal roots.
5. Find $k$ such that $x^2 - 2(k+1)x + k^2 = 0$ has equal roots.
6. If $\alpha, \beta$ are roots of $x^2 - 3x + 1 = 0$, find $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.
7. Find a quadratic with roots $\dfrac{1}{\alpha}, \dfrac{1}{\beta}$ where $\alpha, \beta$ are roots of $x^2 - 5x + 6 = 0$.
8. Solve $x^4 - 5x^2 + 4 = 0$.
9. If one root of $x^2 + bx + c = 0$ is $2 + 3i$ and $b, c$ are real, find $b, c$.
10. Solve $\sqrt{2} x^2 + x + \sqrt{2} = 0$.
11. Find $k$ such that the roots of $x^2 + (k - 1)x + k + 1 = 0$ are real.
12. Show that if the roots of $ax^2 + bx + c = 0$ are reciprocals of each other, then $a = c$.

Pitfalls / Tricks

• Always state the nature of the roots before writing them: discriminant first.
• Complex roots of a real quadratic come in conjugate pairs. If one root is given as $2 + 3i$, the other must be $2 - 3i$.
• Vieta's formulas work even when roots are complex.
• Insight. The quadratic formula, with $\sqrt{\cdot}$ interpreted in $\mathbb{C}$, is universal. Class XII's fundamental theorem of algebra extends this: every polynomial of degree $n$ has exactly $n$ complex roots (counted with multiplicity).