A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a \neq 0\). The number of its roots is determined by the discriminant \(D = b^2 - 4ac\), and the roots themselves are found by the formula \(x_{1,2} = \dfrac{-b \pm \sqrt{D}}{2a}\). This page covers everything on the topic: standard form and coefficients, three types of incomplete equations, the discriminant and its three cases, the root formula and its 'even coefficient' variant, the reduced equation and Vieta's theorem, factorization, and biquadratic equations — with step-by-step examples, analysis of common mistakes, and an interactive trainer.
What is a quadratic equation
A quadratic equation is an equation of the form
where \(x\) is the unknown, and \(a\), \(b\), \(c\) are numbers (coefficients), with \(a \neq 0\). This form is called the standard form of a quadratic equation: terms are arranged in descending order of powers on the left, and zero on the right.
The condition \(a \neq 0\) is not a formality. It is the \(ax^2\) term that makes the equation quadratic: if \(a = 0\), we are left with \(bx + c = 0\), which is a standard linear equation with one root. Therefore, in problems with a parameter, the first step is to check if the leading coefficient becomes zero.
An equation can almost always be reduced to standard form: expand brackets, move all terms to the left side, and combine like terms. For example, \(5x^2 = 3x - 2\) becomes \(5x^2 - 3x + 2 = 0\) after moving terms, where \(a = 5\), \(b = -3\), \(c = 2\). The term 'quadratic equation' means exactly the same thing — it's just another name for it.
Full and incomplete quadratic equations
If all three coefficients are non-zero, the equation is called full. If \(b\) or \(c\) (or both) are zero, it's incomplete. There are exactly three types of incomplete equations, and none of them require the discriminant: each can be solved in one or two lines, which is the quickest way to save time on a test.
The key idea for \(ax^2 + bx = 0\) is factoring out the common factor. The product \(x(ax + b)\) is zero if and only if at least one factor is zero, which immediately gives two roots. For \(ax^2 + c = 0\), the idea is different: isolate \(x^2\) and look at the sign of the resulting number — the square of a real number cannot be negative.
| Equation Form | How to Solve |
|---|---|
| ax2 + c = 02x2 − 50 = 0 | Move c to the right side and divide by a: x2 = −c/a. Take the square root, remembering the second sign.x = 5 and x = −5 |
| ax2 + bx = 0x2 + 7x = 0 | Factor out x: x(ax + b) = 0. The product is zero if one of the factors is zero.x = 0 and x = −7 |
| ax2 = 09x2 = 0 | Divide by a and take the square root of zero. There is always exactly one root.x = 0 |
Discriminant: How to Find It and How Many Roots There Are
The discriminant is a number calculated from the coefficients of a full quadratic equation:
The word itself means 'distinguisher': the discriminant doesn't solve anything, it just distinguishes three situations — two roots, one root, or no roots at all. Therefore, the question 'how many roots does the equation have' is answered without finding the roots themselves.
Calculate \(D\) carefully: \(b^2\) is always positive, even if \(b\) itself is negative, and the signs of \(a\) and \(c\) are included in the product \(4ac\) as they are. Consider \(x^2 + 4x - 5 = 0\): here \(a = 1\), \(b = 4\), \(c = -5\), so \(D = 4^2 - 4 \cdot 1 \cdot (-5) = 16 + 20 = 36\). The discriminant is positive, so there will be two roots.
Quadratic Equation Root Formula
When \(D \ge 0\), the roots are found using the main formula of the topic:
The \(\pm\) sign indicates two calculations: one with addition, the other with subtraction. When \(D = 0\), both calculations yield the same number \(x = -\dfrac{b}{2a}\) — this root is called a double root.
Two common places for errors are: the numerator has a minus sign before \(b\) (for \(b = -7\), this becomes \(+7\)), and the denominator is \(2a\), not just \(2\). When \(a = 1\), the difference is not apparent, so the error surfaces precisely when the leading coefficient is not one.
Let's solve \(2x^2 - 9x - 5 = 0\): \(D = (-9)^2 - 4 \cdot 2 \cdot (-5) = 81 + 40 = 121\), \(\sqrt{D} = 11\), so \(x_1 = \dfrac{9 + 11}{4} = 5\) and \(x_2 = \dfrac{9 - 11}{4} = -0.5\).
Even Second Coefficient: Formula with D₁
If \(b\) is an even number, it's convenient to write it as \(b = 2k\) and calculate the 'reduced' discriminant:
The numbers in the calculations become four times smaller (\(D = 4D_1\)), and the sign of \(D_1\) is the same as that of \(D\) — so the conclusion about the number of roots doesn't change.
Example: \(3x^2 + 8x - 3 = 0\). Here \(b = 8\), so \(k = 4\), and \(D_1 = 4^2 - 3 \cdot (-3) = 16 + 9 = 25\), \(\sqrt{D_1} = 5\). Then \(x_1 = \dfrac{-4 + 5}{3} = \dfrac{1}{3}\) and \(x_2 = \dfrac{-4 - 5}{3} = -3\). Using the standard formula would involve \(D = 100\) — the result is the same, but the arithmetic is more extensive.
Reduced Equation and Vieta's Theorem
An equation with \(a = 1\) is called reduced and written as \(x^2 + px + q = 0\). Any quadratic equation can be reduced to this form by dividing by \(a\): for example, \(4x^2 - 12x + 8 = 0\) becomes \(x^2 - 3x + 2 = 0\) after dividing by 4.
The reduced form is valuable because Vieta's theorem applies to it: the sum of the roots is \(-p\), and the product is \(q\). For \(x^2 - 7x + 10 = 0\), we need two numbers that sum to 7 and multiply to 10 — these are 2 and 5, and we didn't need to calculate the discriminant.
Guessing roots using Vieta's theorem is helpful when the roots are integers and small. It also serves as a free check for answers found using the discriminant: the sum should equal \(-p\), and the product should equal \(q\). Formulas for the general form, the inverse theorem, and the guessing algorithm are discussed in a separate material — Vieta's Theorem.
Factoring a Quadratic Trinomial
If the roots \(x_1\) and \(x_2\) are found, the quadratic trinomial can be factored as:
The factor \(a\) must not be forgotten — without it, the equality is incorrect. Factoring can always be checked easily: expand the brackets and compare with the original trinomial. For instance, the roots of \(x^2 + 2x - 8\) are 2 and \(-4\), so \(x^2 + 2x - 8 = (x - 2)(x + 4)\).
The reverse process also works: if the trinomial can be factored mentally, the roots are immediately apparent, and the discriminant is not needed. Formulas for the difference of squares are helpful here — for example, \(x^2 - 16 = (x - 4)(x + 4)\) gives roots \(4\) and \(-4\) instantly.
A third method to avoid the quadratic formula is completing the square. In \(x^2 + 6x + 5 = 0\), the first two terms can be completed to a square of a sum: \((x + 3)^2 - 9 + 5 = 0\), which means \((x + 3)^2 = 4\), so \(x + 3 = \pm 2\) and \(x = -1\) or \(x = -5\). This is precisely the method used to derive the quadratic formula in its general form.
Equations Reducible to Quadratic: Biquadratic Equations
A biquadratic equation has the form \(ax^4 + bx^2 + c = 0\): only even powers are present. It is solved by substituting \(t = x^2\), with the condition \(t \ge 0\) — the square of a real number cannot be negative.
After the substitution, we get a standard quadratic equation \(at^2 + bt + c = 0\). This is solved using the discriminant, and then for each non-negative root \(t\), we return to \(x = \pm\sqrt{t}\). This explains why there can be up to four roots.
For example, \(x^4 - 5x^2 + 4 = 0\) becomes \(t^2 - 5t + 4 = 0\) after substitution, with roots \(t = 1\) and \(t = 4\). Both are positive, so \(x = \pm 1\) and \(x = \pm 2\) — a total of four roots. If one of the \(t\) values were negative, it would simply be discarded.
This same technique is used to solve rational equations (after bringing them to a common denominator and excluding forbidden values) and equations with substitutions like \(t = x^2 + 3x\).
Examples of Solving Quadratic Equations
Example 1. Full Equation, D > 0
Solve \(3x^2 - 7x + 2 = 0\).
Step 1. The equation is already in standard form: \(a = 3\), \(b = -7\), \(c = 2\).
Step 2. \(D = (-7)^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25 > 0\) — there are two roots.
Step 3. \(\sqrt{D} = 5\), so \(x_{1,2} = \dfrac{7 \pm 5}{6}\): \(x_1 = \dfrac{12}{6} = 2\), \(x_2 = \dfrac{2}{6} = \dfrac{1}{3}\).
Step 4. Check by substitution. \(3 \cdot 2^2 - 7 \cdot 2 + 2 = 12 - 14 + 2 = 0\); \(3 \cdot \left(\tfrac{1}{3}\right)^2 - 7 \cdot \tfrac{1}{3} + 2 = \tfrac{1}{3} - \tfrac{7}{3} + 2 = 0\).
Answer: \(x_1 = 2\), \(x_2 = \dfrac{1}{3}\).
Example 2. Discriminant is Zero
Solve \(4x^2 - 12x + 9 = 0\).
Step 1. \(a = 4\), \(b = -12\), \(c = 9\).
Step 2. \(D = (-12)^2 - 4 \cdot 4 \cdot 9 = 144 - 144 = 0\) — there is one root.
Step 3. \(x = -\dfrac{b}{2a} = \dfrac{12}{8} = 1.5\).
Step 4. Check. \(4 \cdot 1.5^2 - 12 \cdot 1.5 + 9 = 9 - 18 + 9 = 0\).
Answer: \(x = 1.5\). Note: the left side is a perfect square \((2x - 3)^2\), which is why there is only one root.
Example 3. Negative Discriminant
Solve \(2x^2 + 3x + 5 = 0\).
Step 1. \(a = 2\), \(b = 3\), \(c = 5\).
Step 2. \(D = 3^2 - 4 \cdot 2 \cdot 5 = 9 - 40 = -31 < 0\).
Step 3. The discriminant is negative, so \(\sqrt{D}\) cannot be taken among real numbers. The root formula does not need to be applied — the solution is complete.
Answer: No real roots. This is a valid answer, not a sign of an error: the graph of the function \(y = 2x^2 + 3x + 5\) simply does not intersect the x-axis.
Example 4. Incomplete Equation of the form ax² + c = 0
Solve \(3x^2 - 27 = 0\).
Step 1. Move the constant term: \(3x^2 = 27\).
Step 2. Divide by 3: \(x^2 = 9\).
Step 3. The square of two numbers is 9: \(x = 3\) and \(x = -3\). The second sign is most often missed.
Step 4. Check. \(3 \cdot 3^2 - 27 = 0\); \(3 \cdot (-3)^2 - 27 = 27 - 27 = 0\).
Answer: \(x_1 = 3\), \(x_2 = -3\). Compare with the equation \(2x^2 + 8 = 0\): there \(x^2 = -4\), and there are no real roots.
Example 5. Incomplete Equation of the form ax² + bx = 0
Solve \(5x^2 - 15x = 0\).
Step 1. Factor out the common factor: \(5x(x - 3) = 0\).
Step 2. The product is zero if one of the factors is zero: \(5x = 0\) or \(x - 3 = 0\).
Step 3. This gives \(x_1 = 0\), \(x_2 = 3\).
Step 4. Check. \(5 \cdot 0^2 - 15 \cdot 0 = 0\); \(5 \cdot 3^2 - 15 \cdot 3 = 45 - 45 = 0\).
Answer: \(x_1 = 0\), \(x_2 = 3\). If you divide the original equation by \(x\), you are left with only \(x = 3\) — the root \(x = 0\) is lost. You cannot divide by an expression containing a variable.
Example 6. Even Second Coefficient
Solve \(5x^2 - 22x + 8 = 0\).
Step 1. \(b = -22\) is even, so \(k = -11\), and \(a = 5\), \(c = 8\).
Step 2. \(D_1 = k^2 - ac = (-11)^2 - 5 \cdot 8 = 121 - 40 = 81\), \(\sqrt{D_1} = 9\).
Step 3. \(x_{1,2} = \dfrac{-k \pm \sqrt{D_1}}{a} = \dfrac{11 \pm 9}{5}\): \(x_1 = 4\), \(x_2 = \dfrac{2}{5} = 0.4\).
Step 4. Check. \(5 \cdot 4^2 - 22 \cdot 4 + 8 = 80 - 88 + 8 = 0\); \(5 \cdot 0.4^2 - 22 \cdot 0.4 + 8 = 0.8 - 8.8 + 8 = 0\).
Answer: \(x_1 = 4\), \(x_2 = 0.4\). The standard discriminant would give \(D = 324\) — the same answer, but it would take longer to calculate.
Example 7. First, Reduce to Standard Form
Solve \((x - 3)(x + 5) = 9\).
Step 1. Expand the brackets: \(x^2 + 5x - 3x - 15 = 9\), which is \(x^2 + 2x - 15 = 9\).
Step 2. Move the 9 to the left: \(x^2 + 2x - 24 = 0\). Only now can we identify the coefficients: \(a = 1\), \(b = 2\), \(c = -24\).
Step 3. \(D = 2^2 - 4 \cdot 1 \cdot (-24) = 4 + 96 = 100\), \(\sqrt{D} = 10\), \(x_{1,2} = \dfrac{-2 \pm 10}{2}\): \(x_1 = 4\), \(x_2 = -6\).
Step 4. Check in the original equation. \((4 - 3)(4 + 5) = 1 \cdot 9 = 9\); \((-6 - 3)(-6 + 5) = (-9) \cdot (-1) = 9\).
Answer: \(x_1 = 4\), \(x_2 = -6\). A common mistake here is to set each bracket equal to 9: the rule 'the product equals zero' only works for zero.
Example 8. Factoring a Trinomial
Factor the trinomial \(2x^2 - 5x - 3\).
Step 1. Set the trinomial to zero and find the roots: \(D = (-5)^2 - 4 \cdot 2 \cdot (-3) = 25 + 24 = 49\), \(\sqrt{D} = 7\), \(x_{1,2} = \dfrac{5 \pm 7}{4}\): \(x_1 = 3\), \(x_2 = -\dfrac{1}{2}\).
Step 2. Substitute into the formula \(a(x - x_1)(x - x_2)\): \(2(x - 3)\left(x + \tfrac{1}{2}\right)\).
Step 3. It's convenient to distribute the 2 into the second bracket: \((x - 3)(2x + 1)\).
Step 4. Check by expanding. \((x - 3)(2x + 1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3\).
Answer: \(2x^2 - 5x - 3 = (x - 3)(2x + 1)\).
Example 9. Biquadratic Equation
Solve \(x^4 - 13x^2 + 36 = 0\).
Step 1. Substitute \(t = x^2\), where \(t \ge 0\). The equation becomes \(t^2 - 13t + 36 = 0\).
Step 2. \(D = (-13)^2 - 4 \cdot 36 = 169 - 144 = 25\), \(\sqrt{D} = 5\), \(t_{1,2} = \dfrac{13 \pm 5}{2}\): \(t_1 = 9\), \(t_2 = 4\).
Step 3. Both values are non-negative, so we return to \(x\): from \(x^2 = 9\), we get \(x = \pm 3\), and from \(x^2 = 4\), we get \(x = \pm 2\).
Step 4. Check. \(3^4 - 13 \cdot 3^2 + 36 = 81 - 117 + 36 = 0\); \(2^4 - 13 \cdot 2^2 + 36 = 16 - 52 + 36 = 0\). Even powers yield the same results for negative values.
Answer: \(x = \pm 2\) and \(x = \pm 3\) — four roots.
Common Mistakes in Solving Quadratic Equations
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Calculating the discriminant without reducing the equation to standard form: for $2x^2 + 3x = x^2 + 5$, immediately using $a = 2$, $b = 3$, $c = 0$.
First, move all terms to the left and combine like terms: \(2x^2 + 3x - x^2 - 5 = 0\), which is \(x^2 + 3x - 5 = 0\). Only then can we identify \(a = 1\), \(b = 3\), \(c = -5\). The root formula works exclusively with the standard form.
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Losing the sign when squaring $b$: for $x^2 - 9x + 14 = 0$, writing $D = -81 - 56$.
The square of any number is non-negative: \((-9)^2 = 81\), not \(-81\). Correctly, \(D = 81 - 56 = 25\), roots \(x = \dfrac{9 \pm 5}{2}\), which are 7 and 2. Check: \(49 - 63 + 14 = 0\) and \(4 - 18 + 14 = 0\).
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Leaving $b$ instead of $-b$ in the numerator: for $x^2 + 5x - 6 = 0$, calculating $x = \dfrac{5 \pm 7}{2}$.
The formula has \(-b\), so when \(b = 5\), the numerator is \(-5\): \(x = \dfrac{-5 \pm 7}{2}\), giving \(x_1 = 1\), \(x_2 = -6\). Substitution confirms: \(1 + 5 - 6 = 0\) and \(36 - 30 - 6 = 0\).
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Writing 'no roots' when $D = 0$: for $x^2 + 14x + 49 = 0$, the discriminant is zero, and the solution is stopped.
There are no roots only when \(D < 0\). When \(D = 0\), there is exactly one root: \(x = -\dfrac{b}{2a} = -\dfrac{14}{2} = -7\). Check: \(49 - 98 + 49 = 0\).
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Dividing the equation $4x^2 - 9x = 0$ by $x$ and getting only one root $x = 2.25$.
You cannot divide by an expression containing a variable: \(x\) could be zero, and the root \(x = 0\) would be lost. Correctly, factor out \(x\): \(x(4x - 9) = 0\), giving \(x_1 = 0\) and \(x_2 = 2.25\). Check: \(4 \cdot 2.25^2 - 9 \cdot 2.25 = 20.25 - 20.25 = 0\).
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Writing only $x = 4$ for the equation $x^2 = 16$.
The square of \(-4\) is also 16. Both roots must be written: \(x = \pm 4\). It's more reliable to solve using the standard form \(x^2 - 16 = 0\), which is \((x - 4)(x + 4) = 0\) — then both factors are immediately visible.
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Not checking the condition $a \neq 0$ in an equation with a parameter: for $(m - 2)x^2 + 3x - 1 = 0$, immediately calculating the discriminant.
When \(m = 2\), the leading coefficient becomes zero, and the equation becomes linear \(3x - 1 = 0\) with a single root \(x = \dfrac{1}{3}\). The case \(m = 2\) is considered separately, and the root formula is applied only when \(m \neq 2\).
Questions and Answers
What is a quadratic equation?
It's an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), \(c\) are numbers and \(a \neq 0\). The unknown is squared, so there can be two, one, or no roots.
What is the standard form of a quadratic equation?
The standard (general) form is \(ax^2 + bx + c = 0\): all terms are collected on the left in descending order of powers, with zero on the right. Any quadratic equation is reduced to this form before calculating the discriminant.
How do you find the discriminant?
Using the formula \(D = b^2 - 4ac\), substituting the coefficients along with their signs. For example, for \(x^2 + 3x - 10 = 0\), we get \(D = 3^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49\), and the roots are \(\dfrac{-3 \pm 7}{2}\), which are 2 and \(-5\).
What is the formula for the roots of a quadratic equation?
\(x_{1,2} = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), or more concisely \(x_{1,2} = \dfrac{-b \pm \sqrt{D}}{2a}\). It is used when \(D \ge 0\); the denominator is specifically \(2a\), not just 2.
How many roots can a quadratic equation have?
At most two. If \(D > 0\), there are two roots; if \(D = 0\), there is one root (a double root); if \(D < 0\), there are no real roots.
What is an incomplete quadratic equation?
It's an equation where \(b = 0\), \(c = 0\), or both: \(ax^2 + c = 0\), \(ax^2 + bx = 0\), \(ax^2 = 0\). All three are solved without the discriminant — by moving terms or factoring out \(x\).
How to solve quadratic equations without the discriminant?
There are three ways: guessing roots using Vieta's theorem for a reduced equation, factoring, and completing the square. Guessing works fastest when the roots are small integers.
What to do if the discriminant is negative?
State that there are no real roots and finish the solution. A negative discriminant is a normal answer: the graph of the corresponding parabola does not intersect the x-axis. Roots in such cases exist only among complex numbers, which are studied later.
How to check the found roots?
Substitute each root back into the original equation — it should result in a true statement. For a reduced equation, there's also a quick check using Vieta's theorem: the sum of the roots equals \(-p\), and the product equals \(q\).
Are quadratic and quadratic equations the same thing?
Yes, these are two names for the same object \(ax^2 + bx + c = 0\). Don't confuse it with a quadratic function \(y = ax^2 + bx + c\): the roots of the equation are precisely the x-coordinates of the points where the function's graph intersects the x-axis.
In which grade are quadratic equations taught?
In 8th grade algebra: first incomplete equations, then the discriminant, the root formula, and Vieta's theorem. The topic is further explored with quadratic functions, inequalities, and problems in the OGE and ЕГЭ exams.
