What Is The Nature Of Roots?
The nature of roots of a quadratic equation is the type of solutions it has, real or complex, equal or distinct, and rational or irrational, read off before the roots are computed. Every quadratic equation $ax^2 + bx + c = 0$ (with $a \neq 0$) has exactly two roots, and one number decides what those two roots look like.
That number is the discriminant, $D = b^2 - 4ac$. The rest of this page is worked examples: enough of them, across every case, that the pattern becomes automatic. The actual roots can wait until the type is known.
How Do You Find The Nature Of Roots From The Discriminant?
Three steps run under every example below.
Rearrange the equation into standard form $ax^2 + bx + c = 0$, then read off $a$, $b$, and $c$ with their signs.
Compute the discriminant $D = b^2 - 4ac$.
Match the sign of $D$ to the table, and when $D > 0$ run one extra perfect-square check.
Table 1: How the discriminant sign fixes the nature of the roots.
Discriminant $D = b^2 - 4ac$ | Nature of the roots | Graph vs the x-axis |
|---|---|---|
$D > 0$ | Two distinct real roots | Crosses at two points |
$D = 0$ | One real repeated root | Touches at one point |
$D < 0$ | Two complex conjugate roots | Never meets the axis |
When $D > 0$, a second test splits the real roots further.
Table 2: For $D > 0$, the perfect-square check splits rational from irrational.
Discriminant with $D > 0$ | Nature of the real roots |
|---|---|
$D$ is a perfect square ($1, 4, 9, 16, 25, \dots$) | Two distinct rational roots |
$D$ is not a perfect square | Two distinct irrational roots |
What Do Two Distinct Rational Roots Look Like? (D > 0, Perfect Square)
When $D$ is positive and a perfect square, its square root is a whole number, so the quadratic formula returns clean fractions or integers. These are the roots that also factor neatly over the rationals.
Example 1: Find the nature of roots of $3x^2 - 10x + 3 = 0$.
Read the coefficients: $a = 3$, $b = -10$, $c = 3$.
$$D = (-10)^2 - 4(3)(3) = 100 - 36 = 64$$
Since $D = 64 > 0$ and $64 = 8^2$ is a perfect square, the roots are two distinct rational numbers:
$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{10 \pm 8}{6} = 3 \ \text{or} \ \tfrac{1}{3}$$
Final answer: two distinct rational roots, $x = 3$ and $x = \tfrac{1}{3}$.
Example 2: Find the nature of roots of $6x^2 + x - 2 = 0$.
Here $a = 6$, $b = 1$, $c = -2$. Watch the negative $c$:
$$D = (1)^2 - 4(6)(-2) = 1 + 48 = 49$$
$D = 49 = 7^2$, a perfect square, so again two distinct rational roots:
$$x = \frac{-1 \pm 7}{12} = \tfrac{1}{2} \ \text{or} \ -\tfrac{2}{3}$$
A perfect-square discriminant is also a signal that factorization will work, which is why these two equations factor without surds.
What Do Two Distinct Irrational Roots Look Like? (D > 0, Not A Perfect Square)
When $D$ is positive but not a perfect square, its square root does not simplify to a whole number, so the roots are irrational surds that come as a conjugate pair.
Example 3: Find the nature of roots of $2x^2 - 8x + 3 = 0$.
With $a = 2$, $b = -8$, $c = 3$:
$$D = (-8)^2 - 4(2)(3) = 64 - 24 = 40$$
$D = 40 > 0$, and $40$ is not a perfect square (it sits between $36 = 6^2$ and $49 = 7^2$), so the roots are two distinct irrational numbers:
$$x = \frac{8 \pm \sqrt{40}}{4} = \frac{8 \pm 2\sqrt{10}}{4} = 2 \pm \frac{\sqrt{10}}{2}$$
Final answer: two distinct irrational roots, $x = 2 \pm \dfrac{\sqrt{10}}{2}$.
Example 4: Find the nature of roots of $x^2 - 6x + 7 = 0$.
Here $a = 1$, $b = -6$, $c = 7$:
$$D = (-6)^2 - 4(1)(7) = 36 - 28 = 8$$
$D = 8 > 0$ and is not a perfect square, so the roots are irrational conjugates:
$$x = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}$$
Notice both irrational examples still have $D > 0$, so both graphs cross the x-axis twice. The perfect-square check changes the flavour of the roots, not how many there are.
What Does A Repeated Root Look Like? (D = 0)
When $D = 0$, the $\pm$ in the quadratic formula adds and subtracts zero, so the two roots collapse into one value. This is one repeated root, not zero roots.
Example 5: Find the nature of roots of $9x^2 - 12x + 4 = 0$.
With $a = 9$, $b = -12$, $c = 4$:
$$D = (-12)^2 - 4(9)(4) = 144 - 144 = 0$$
Since $D = 0$, there is exactly one real repeated root at $x = -\dfrac{b}{2a}$:
$$x = -\frac{-12}{2(9)} = \frac{12}{18} = \frac{2}{3}$$
Final answer: one repeated real root, $x = \tfrac{2}{3}$ (the equation is $(3x - 2)^2 = 0$).
Example 6: Find the nature of roots of $x^2 + 14x + 49 = 0$.
Here $a = 1$, $b = 14$, $c = 49$:
$$D = (14)^2 - 4(1)(49) = 196 - 196 = 0$$
$D = 0$, so one repeated real root:
$$x = -\frac{14}{2(1)} = -7$$
Both $D = 0$ equations are perfect-square trinomials, which is exactly what a zero discriminant detects.
What Do Complex Roots Look Like? (D < 0)
When $D < 0$, the square root is of a negative number, which is not real. The roots leave the number line and arrive as a pair of complex conjugates.
Example 7: Find the nature of roots of $x^2 - 2x + 5 = 0$.
With $a = 1$, $b = -2$, $c = 5$:
$$D = (-2)^2 - 4(1)(5) = 4 - 20 = -16$$
Since $D = -16 < 0$, the roots are two complex conjugates:
$$x = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i$$
Final answer: two complex conjugate roots, $x = 1 + 2i$ and $x = 1 - 2i$.
Example 8: Find the nature of roots of $2x^2 + 3x + 5 = 0$.
Here $a = 2$, $b = 3$, $c = 5$:
$$D = (3)^2 - 4(2)(5) = 9 - 40 = -31$$
$D = -31 < 0$, so two complex conjugate roots:
$$x = \frac{-3 \pm \sqrt{-31}}{4} = \frac{-3 \pm i\sqrt{31}}{4}$$
The two roots are mirror images across the real axis, so the size of each is the same modulus. A negative discriminant is the only case that produces no real roots at all.
How Do You Find k So The Roots Are Equal?
A very common exam version hides a letter in the coefficients and asks for the value that forces a chosen nature of roots. Equal roots means $D = 0$, so you set the discriminant to zero and solve for the unknown.
Example 9: Find the value of $k$ for which $4x^2 + kx + 9 = 0$ has equal roots.
Equal roots require $D = 0$. Read $a = 4$, $b = k$, $c = 9$, then set the discriminant to zero:
$$D = k^2 - 4(4)(9) = 0$$
$$k^2 - 144 = 0$$
$$k^2 = 144$$
$$k = \pm 12$$
Final answer: the roots are equal when $k = 12$ or $k = -12$. Both values make the left side a perfect-square trinomial, $(2x \pm 3)^2$.
The same method finds the condition for other cases: use $D > 0$ for two distinct real roots and $D < 0$ for complex roots, which turns the equation into an inequality in the unknown.
A Word Problem Example On The Nature Of Roots
The discriminant answers a real question: does a solution even exist? A negative discriminant means the situation being modelled is impossible.
Example 10: A gardener has $40$ metres of fencing to enclose a rectangular bed along a straight wall, using the wall as one long side. Can the bed enclose an area of $250$ square metres?
Let the width (each of the two sides running out from the wall) be $x$ metres. The fencing covers two widths and one length, so the length is $40 - 2x$, and the area is:
$$x(40 - 2x) = 250$$
Rearrange into standard form:
$$40x - 2x^2 = 250 \implies 2x^2 - 40x + 250 = 0 \implies x^2 - 20x + 125 = 0$$
Now read the nature of roots. With $a = 1$, $b = -20$, $c = 125$:
$$D = (-20)^2 - 4(1)(125) = 400 - 500 = -100$$
$D = -100 < 0$, so there is no real value of $x$. Final answer: the bed cannot enclose $250$ square metres with $40$ metres of fencing. If the target were $150$ square metres instead, the equation becomes $x^2 - 20x + 75 = 0$ with $D = 400 - 300 = 100 > 0$, a perfect square, giving two real widths ($x = 5$ or $x = 15$) that both work.
Nature Of Roots Examples At A Glance
Every worked equation above, side by side, so the pattern from discriminant to root type is visible in one view.
Table 3: Summary of the nature of roots examples on this page.
Example | Equation | $D = b^2 - 4ac$ | Nature of the roots |
|---|---|---|---|
1 | $3x^2 - 10x + 3 = 0$ | $64$ | Two distinct rational |
2 | $6x^2 + x - 2 = 0$ | $49$ | Two distinct rational |
3 | $2x^2 - 8x + 3 = 0$ | $40$ | Two distinct irrational |
4 | $x^2 - 6x + 7 = 0$ | $8$ | Two distinct irrational |
5 | $9x^2 - 12x + 4 = 0$ | $0$ | One repeated real |
6 | $x^2 + 14x + 49 = 0$ | $0$ | One repeated real |
7 | $x^2 - 2x + 5 = 0$ | $-16$ | Two complex conjugates |
8 | $2x^2 + 3x + 5 = 0$ | $-31$ | Two complex conjugates |
Read down the discriminant column and the rule reads itself: positive splits into rational or irrational by the perfect-square test, zero gives a repeated root, negative gives complex conjugates.
Why Does The Discriminant Decide The Nature Of Roots?
The discriminant is not a separate rule to memorise. It is the part of the quadratic formula that sits under the square root, and once you see it there, all the examples above explain themselves.
Every root comes from the formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{D}}{2a}$$
The only piece that changes character from equation to equation is $\sqrt{D}$.
If $D > 0$, then $\sqrt{D}$ is a real positive number, so the $\pm$ gives two different real answers. It is rational when $\sqrt{D}$ is a whole number (a perfect square) and irrational otherwise.
If $D = 0$, then $\sqrt{D} = 0$, so adding and subtracting zero give the same value, and the two roots fall onto one repeated root, $x = -\tfrac{b}{2a}$.
If $D < 0$, then $\sqrt{D}$ is the square root of a negative number, which is not real, so the $\pm$ produces a matched pair of complex conjugates.
The graph tells the same story. The curve $y = ax^2 + bx + c$ is a parabola, and its real roots are the points where the parabola meets the x-axis: two crossings, one touch, or none. That formula itself is built by completing the square, so the discriminant is a fingerprint left behind by that derivation.
Who Discovered The Nature Of Roots?
People were solving quadratics more than a thousand years before anyone wrote $b^2 - 4ac$. The rule travelled through India and the Islamic world as a recipe in words, and the label "discriminant" arrived only in the 1800s.
Two more mathematicians shaped how we describe roots today:
Sridhara (around 870–930 CE, India) wrote an early clear rule for solving a quadratic, the method still called the Sridharacharya rule. His own book was lost, and the rule survives because a later mathematician quoted it centuries afterward.
James Joseph Sylvester (1814–1897, England) coined the word "discriminant" in 1851, naming the quantity $b^2 - 4ac$ that every example on this page turns on.
Where Are Nature Of Roots Examples Used In The Real World?
Checking a discriminant asks "does a real answer exist, and is it unique?", a question that shows up well outside the exam hall.
Projectile motion: whether a thrown or launched object reaches a target height is whether its quadratic height equation has real roots, exactly like the high-jump image and the fencing example. $D \geq 0$ means it gets there; $D < 0$ means it falls short.
Engineering and control systems: a suspension or a circuit has a characteristic quadratic whose discriminant sets its behaviour, $D > 0$ overdamped, $D = 0$ critically damped, $D < 0$ underdamped and oscillating.
Computer graphics: testing whether a ray of light hits a sphere reduces to a quadratic, and the sign of the discriminant tells the renderer whether the ray misses, grazes, or passes through.
Economics: profit and cost models are often quadratic, and real roots mark the break-even quantities where profit crosses zero.
One test on three coefficients decides "possible or not" across flight, machinery, rendering, and money, fields that look nothing alike.
What Are The Most Common Nature Of Roots Mistakes?
These four errors cost the most marks on discriminant questions, matching the sign, standard-form, and perfect-square slips flagged across student-error guides.
Losing a sign inside $b^2 - 4ac$.
Where it slips in:
With a negative $b$ or a negative $c$, a student writes $b^2 + 4ac$, or accidentally turns $-4ac$ positive.
Don't do this:
Do not drop the signs of $b$ or $c$. Note that $b^2$ is never negative, even when $b$ is.
The correct way:
Compute $b^2$ first (always non-negative), then subtract $4ac$ with the real signs. For Example 2, $a = 6$, $b = 1$, $c = -2$ gives $D = 1 - 4(6)(-2) = 1 + 48 = 49$, not $1 - 48$.
Reading $D = 0$ as "no roots."
Where it slips in:
A student sees the discriminant is not positive and concludes the equation has no solution.
Don't do this:
Do not read $D = 0$ as an empty answer. Only $D < 0$ gives no real roots.
The correct way:
Read $D = 0$ as one real repeated root at $x = -\tfrac{b}{2a}$, as in Examples 5 and 6, where the two roots land on the same value.
Skipping the perfect-square check.
Where it slips in:
A student finds $D > 0$, writes "two real roots," and stops, missing whether those roots are rational or irrational.
Don't do this:
Do not assume every positive discriminant gives clean, rational answers.
The correct way:
When $D > 0$, test whether it is a perfect square. A perfect square gives rational roots (Examples 1 and 2); anything else gives irrational conjugate surds (Examples 3 and 4).
Reading coefficients before writing standard form.
Where it slips in:
The equation arrives as $x(40 - 2x) = 250$ or $3x^2 = 10x - 3$, and a student reads $a$, $b$, $c$ straight off it.
Don't do this:
Do not pull coefficients from an equation that is not yet set equal to zero.
The correct way:
Move every term to one side first, into $ax^2 + bx + c = 0$, exactly as the word-problem example did before computing $D$.
Practice Problems On Nature Of Roots
For each equation, compute the discriminant and state the nature of its roots. Answers follow each line.
$x^2 - 8x + 12 = 0$.
(Answer: $D = 64 - 48 = 16 = 4^2$, so two distinct rational roots.)$5x^2 - 2x + 1 = 0$.
(Answer: $D = 4 - 20 = -16 < 0$, so two complex conjugate roots.)$x^2 - 4x - 2 = 0$.
(Answer: $D = 16 + 8 = 24$, positive but not a perfect square, so two distinct irrational roots.)$16x^2 - 24x + 9 = 0$.
(Answer: $D = 576 - 576 = 0$, so one repeated real root, $x = \tfrac{3}{4}$.)$3x^2 + 5x - 2 = 0$.
(Answer: $D = 25 + 24 = 49 = 7^2$, so two distinct rational roots.)Find $m$ so that $x^2 - mx + 16 = 0$ has equal roots.
(Answer: equal roots need $D = 0$, so $m^2 - 64 = 0$ and $m = \pm 8$.)
Where Should You Go Next After Nature Of Roots Examples?
Working these examples opens straight into finding and using the roots themselves.
Roots Of Quadratic Equation. Move from "what type of roots" to computing the exact values with the quadratic formula.
Completing The Square. See the method the quadratic formula and the discriminant are both built from.
Factorization Of Quadratic Polynomials. When the discriminant is a perfect square, the equation factors over the rationals, and this is how.
If your child is drilling these examples, a live Bhanzu trainer teaches the nature of roots starting from the parabola and the square-root term, so each case is understood rather than memorised, in the Bhanzu algebra program.
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