A quadratic equation has the form $ax^2 + bx + c = 0$ with $a \ne 0$. Its roots are given by the quadratic formula, and the quantity under the square root — the discriminant $D = b^2 - 4ac$ — decides the nature of those roots.
$$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Over the real numbers, a negative discriminant means "no solution". Once complex numbers exist, that gap closes: a negative $D$ simply produces a square root of a negative number, which we write using $i$. The result is a pair of complex roots in the form $a + ib$.
When $D < 0$, write $\sqrt{D} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2}$, so the roots become:
$$x = \dfrac{-b \pm i\sqrt{4ac - b^2}}{2a}$$
For an equation with real coefficients, complex roots always arrive in conjugate pairs: if $p + iq$ is a root, then $p - iq$ is the other. This is why a quadratic over the reals can never have exactly one non-real root. (If the coefficients are themselves complex, this pairing need not hold, and you simply apply the quadratic formula directly.)
Relations between roots and coefficients still hold, and they hold even when the roots are complex:
$$\alpha + \beta = -\dfrac{b}{a}, \qquad \alpha\beta = \dfrac{c}{a}$$
These let you check an answer quickly: for conjugate roots $p \pm iq$, the sum is the real number $2p$ and the product is $p^2 + q^2$, both real, exactly as $-b/a$ and $c/a$ must be.
Building a quadratic from its roots. The same relations run in reverse. If you know the two roots $\alpha$ and $\beta$, the quadratic (with leading coefficient $1$) is:
$$x^2 - (\alpha + \beta)\,x + \alpha\beta = 0$$
So if real coefficients are required and one root is $p + iq$, the other must be $p - iq$; then sum $= 2p$ and product $= p^2 + q^2$ give the equation $x^2 - 2px + (p^2 + q^2) = 0$ at once.
Deeper Insight — why complex roots are a feature, not a failure: A negative discriminant used to mark the end of the road, but it is really a signpost that the answers live one dimension over, off the real line. The reason this works so cleanly is the conjugate pairing forced by real coefficients: the imaginary parts of $p + iq$ and $p - iq$ cancel in the sum and combine into the real quantity $p^2 + q^2$ in the product, so the coefficients $-b/a$ and $c/a$ stay perfectly real even though the roots are not. This is a small instance of the Fundamental Theorem of Algebra, which promises that a degree-$n$ polynomial has exactly $n$ roots in the complex numbers, counted with multiplicity — no equation is ever truly unsolvable. The practical lesson is to stop treating $D < 0$ as an error and start reading it as an instruction: factor out the $i$, simplify the surd, and report the conjugate pair.
Solve $x^2 + 1 = 0$.
Solution- Rearrange: $x^2 = -1$.
- Take square roots: $x = \pm\sqrt{-1} = \pm i$.
Answer: $x = i$ or $x = -i$
Solve $x^2 + 4 = 0$.
Solution- Rearrange: $x^2 = -4$.
- $x = \pm\sqrt{-4} = \pm\sqrt{4}\,i = \pm 2i$.
Answer: $x = 2i$ or $x = -2i$
Solve $x^2 - 2x + 5 = 0$.
Solution- Here $a = 1,\ b = -2,\ c = 5$, so $D = (-2)^2 - 4(1)(5) = 4 - 20 = -16$.
- $\sqrt{D} = \sqrt{-16} = 4i$.
- $x = \dfrac{-(-2) \pm 4i}{2(1)} = \dfrac{2 \pm 4i}{2}$.
- Simplify: $x = 1 \pm 2i$.
Answer: $x = 1 + 2i$ or $x = 1 - 2i$
Solve $2x^2 + x + 1 = 0$.
Solution- $a = 2,\ b = 1,\ c = 1$, so $D = 1^2 - 4(2)(1) = 1 - 8 = -7$.
- $\sqrt{D} = \sqrt{-7} = i\sqrt{7}$.
- $x = \dfrac{-1 \pm i\sqrt{7}}{2(2)} = \dfrac{-1 \pm i\sqrt{7}}{4}$.
Answer: $x = -\dfrac{1}{4} \pm \dfrac{\sqrt{7}}{4}i$
Solve $x^2 - 4x + 13 = 0$ and verify the sum and product of its roots.
Solution- $a = 1,\ b = -4,\ c = 13$, so $D = 16 - 52 = -36$ and $\sqrt{D} = 6i$.
- $x = \dfrac{4 \pm 6i}{2} = 2 \pm 3i$.
- Sum of roots: $(2 + 3i) + (2 - 3i) = 4 = -\dfrac{b}{a} = -\dfrac{-4}{1}$. ✓
- Product: $(2 + 3i)(2 - 3i) = 4 - 9i^2 = 4 + 9 = 13 = \dfrac{c}{a}$. ✓
Answer: $x = 2 \pm 3i$; sum $= 4$, product $= 13$, both verified.
Solve $\sqrt{3}\,x^2 - \sqrt{2}\,x + 3\sqrt{3} = 0$.
Solution- $a = \sqrt{3},\ b = -\sqrt{2},\ c = 3\sqrt{3}$.
- $D = (-\sqrt{2})^2 - 4(\sqrt{3})(3\sqrt{3}) = 2 - 36 = -34$.
- $\sqrt{D} = i\sqrt{34}$, and $2a = 2\sqrt{3}$.
- $x = \dfrac{\sqrt{2} \pm i\sqrt{34}}{2\sqrt{3}}$.
Answer: $x = \dfrac{\sqrt{2} \pm i\sqrt{34}}{2\sqrt{3}}$
Solve $x^2 + 3x + 5 = 0$.
Solution- $a = 1,\ b = 3,\ c = 5$, so $D = 3^2 - 4(1)(5) = 9 - 20 = -11$.
- $\sqrt{D} = i\sqrt{11}$.
- $x = \dfrac{-3 \pm i\sqrt{11}}{2}$.
Answer: $x = -\dfrac{3}{2} \pm \dfrac{\sqrt{11}}{2}i$
Form a quadratic equation with real coefficients, one of whose roots is $3 - 2i$.
Solution- Real coefficients force the conjugate $3 + 2i$ to be the other root.
- Sum of roots: $(3 - 2i) + (3 + 2i) = 6$.
- Product of roots: $(3 - 2i)(3 + 2i) = 9 - 4i^2 = 9 + 4 = 13$.
- Use $x^2 - (\text{sum})x + (\text{product}) = 0$.
Answer: $x^2 - 6x + 13 = 0$
Solve $x^2 - x + 1 = 0$ over the complex numbers.
Solution- $a = 1,\ b = -1,\ c = 1$, so $D = (-1)^2 - 4(1)(1) = 1 - 4 = -3$.
- $\sqrt{D} = i\sqrt{3}$.
- $x = \dfrac{1 \pm i\sqrt{3}}{2}$.
Answer: $x = \dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i$
If one root of $x^2 + px + q = 0$ (with real $p, q$) is $2 + i$, find $p$ and $q$.
Solution- Real coefficients mean the other root is the conjugate $2 - i$.
- Sum of roots $= 4 = -p$, so $p = -4$.
- Product of roots $= (2 + i)(2 - i) = 4 - i^2 = 5 = q$.
Answer: $p = -4,\ q = 5$, giving $x^2 - 4x + 5 = 0$.
Solve the equation $x^2 + x + \dfrac{1}{\sqrt{2}} = 0$.
Solution- $a = 1,\ b = 1,\ c = \dfrac{1}{\sqrt{2}}$, so $D = 1 - 4\cdot\dfrac{1}{\sqrt{2}} = 1 - \dfrac{4}{\sqrt{2}} = 1 - 2\sqrt{2}$.
- Since $2\sqrt{2} \approx 2.83 > 1$, $D = 1 - 2\sqrt{2} < 0$, so $\sqrt{D} = i\sqrt{2\sqrt{2} - 1}$.
- $x = \dfrac{-1 \pm i\sqrt{2\sqrt{2} - 1}}{2}$.
Answer: $x = \dfrac{-1 \pm i\sqrt{2\sqrt{2} - 1}}{2}$
Solve $4x^2 + 1 = 0$ and state the conjugate-pair structure of the roots.
Solution- Rearrange: $x^2 = -\dfrac{1}{4}$.
- $x = \pm\sqrt{-\dfrac{1}{4}} = \pm\dfrac{1}{2}i$.
- The roots $\dfrac{1}{2}i$ and $-\dfrac{1}{2}i$ are conjugates ($p = 0,\ q = \tfrac{1}{2}$), consistent with real coefficients.
Answer: $x = \pm\dfrac{1}{2}i$ — a conjugate pair.