What Is The Sridharacharya Formula?
The Sridharacharya formula gives the exact solutions of any quadratic equation written in standard form, $ax^2 + bx + c = 0$, as long as $a \neq 0$. Known worldwide as the quadratic formula, it is:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
To use it, read off the three numbers $a$, $b$, and $c$ from the equation, substitute them, and simplify. The $\pm$ sign is doing real work: it splits into two calculations, one with a plus and one with a minus, which is why a quadratic usually has two solutions.
Table: The parts of the Sridharacharya formula, and what each symbol means.
Symbol | Meaning |
|---|---|
$a$ | Coefficient of $x^2$, the quadratic term. Must be non-zero. |
$b$ | Coefficient of $x$, the linear term. |
$c$ | The constant term (no $x$ attached). |
$x$ | The unknown you are solving for; its values are the roots. |
$\pm$ | Splits the formula into two roots, one using $+$ and one using $-$. |
$b^2 - 4ac$ | The discriminant, $D$. Its sign fixes how many real roots exist. |
The roots are the values of $x$ that make the equation true. Graphically, they are the points where the curve $y = ax^2 + bx + c$ meets the x-axis.
Where Does The Sridharacharya Formula Come From?
The formula is completing the square carried out once, in general, with the letters left in place of numbers. Sridharacharya's own method avoids fractions by multiplying through by $4a$ first, and that is the route shown here.
Start from the standard form.
$$ax^2 + bx + c = 0$$
Multiply every term by $4a$. This clears the way to a clean square later.
$$4a^2x^2 + 4abx + 4ac = 0$$
Move the constant term to the right-hand side.
$$4a^2x^2 + 4abx = -4ac$$
Add $b^2$ to both sides. This is the step that turns the left side into a perfect square.
$$4a^2x^2 + 4abx + b^2 = b^2 - 4ac$$
The left side now folds into a single squared bracket.
$$(2ax + b)^2 = b^2 - 4ac$$
Take the square root of both sides. The $\pm$ enters here, because a square root can be positive or negative.
$$2ax + b = \pm\sqrt{b^2 - 4ac}$$
Subtract $b$, then divide both sides by $2a$.
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
That is the Sridharacharya formula, derived line by line from the equation it solves.
How Do You Use The Sridharacharya Formula?
To solve a quadratic, match it to $ax^2 + bx + c = 0$, list $a$, $b$, and $c$ with their signs, then substitute into the formula. The signs are where most marks are lost, so write them down before touching the formula.
Example 1: Solve $2x^2 - x - 3 = 0$.
First identify the coefficients. Comparing with $ax^2 + bx + c = 0$ gives $a = 2$, $b = -1$, and $c = -3$.
Compute the discriminant.
$$b^2 - 4ac = (-1)^2 - 4(2)(-3) = 1 + 24 = 25$$
Since $25 > 0$, there are two distinct real roots. Now substitute into the formula, taking care that $-b = -(-1) = +1$.
$$x = \frac{-(-1) \pm \sqrt{25}}{2(2)} = \frac{1 \pm 5}{4}$$
Split the $\pm$ into the two roots.
$$x = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2}, \qquad x = \frac{1 - 5}{4} = \frac{-4}{4} = -1$$
Final answer: $x = \frac{3}{2}$ or $x = -1$. You can check either root by substituting it back: $2\left(\tfrac{3}{2}\right)^2 - \tfrac{3}{2} - 3 = \tfrac{9}{2} - \tfrac{3}{2} - 3 = 0$.
What Does The Discriminant Tell You About The Roots?
The discriminant $D = b^2 - 4ac$ is the part of the formula under the square root, and its sign settles the nature of the roots before you finish solving. Reading it first tells you what kind of answer to expect.
Table: How the discriminant decides the nature of the roots.
Discriminant $D = b^2 - 4ac$ | Nature of the roots |
|---|---|
$D > 0$ | Two distinct real roots |
$D = 0$ | Two equal real roots (one repeated root) |
$D < 0$ | No real roots (two complex conjugate roots) |
When $D = 0$, the square root vanishes and the formula collapses to $x = \frac{-b}{2a}$; the equation $x^2 - 6x + 9 = 0$ is one such case, with $x = 3$ counted twice. When $D < 0$, the square root has no real value, so the roots of the quadratic equation are a pair of complex numbers rather than points on the x-axis.
Example 2: Solve $x^2 - 4x + 1 = 0$.
Here $a = 1$, $b = -4$, and $c = 1$, so the discriminant is $(-4)^2 - 4(1)(1) = 16 - 4 = 12$. That is positive, so expect two real roots, and because $12$ is not a perfect square, they will be irrational.
$$x = \frac{-(-4) \pm \sqrt{12}}{2(1)} = \frac{4 \pm \sqrt{12}}{2}$$
Simplify the surd: $\sqrt{12} = 2\sqrt{3}$.
$$x = \frac{4 \pm 2\sqrt{3}}{2}$$
Divide the whole numerator, both terms, by $2$.
$$x = 2 \pm \sqrt{3}$$
Final answer: $x = 2 + \sqrt{3}$ or $x = 2 - \sqrt{3}$.
Why Does The Sridharacharya Formula Always Work?
Factoring only lands when the roots happen to be tidy whole numbers or simple fractions. The Sridharacharya formula has no such limit, and there are three reasons worth naming.
It is completing the square, generalised. Every quadratic can be rewritten as a perfect square plus a constant, so every quadratic can be solved by the same square-root step. The formula just does that rearrangement once, with letters, so you never have to repeat it.
The $\pm$ is the parabola's symmetry. A parabola is a mirror-symmetric curve, and its two roots sit at equal distances on either side of the line $x = \frac{-b}{2a}$. The plus and minus in the formula are that mirror, measured out by $\frac{\sqrt{b^2 - 4ac}}{2a}$.
The discriminant is the switch. Whether those two mirror points land on the x-axis, meet at it, or miss it entirely depends only on the sign of $b^2 - 4ac$.
So the formula is not a lucky trick to memorise. It is what completing the square becomes when you carry it out for every quadratic at the same time.
Who Invented The Sridharacharya Formula?
The method behind the formula was written down in India more than a thousand years ago, long before algebra used the symbols we write today. The person it is named for very nearly slipped out of the record entirely.
Three Indian mathematicians shaped this corner of algebra:
Sridharacharya (c. 870–930 CE, India) wrote the practical mathematics texts Patiganita and Trisatika, and is remembered as one of the first to give a general rule for the quadratic.
Bhaskara II (1114–1185 CE, India), also called Bhaskaracharya, preserved Sridhara's rule and showed clearly that a single quadratic can have more than one solution.
Brahmagupta (598–668 CE, India) had given an even earlier rule for solving quadratics, written entirely in words, centuries before symbolic algebra existed.
Where Is The Sridharacharya Formula Used In The Real World?
Quadratics appear wherever a quantity rises and then turns back, and the formula is how the turning points get found.
Physics and engineering: the height of a thrown ball or a launched rocket is a quadratic in time, so projectile motion problems ask a quadratic question: at what times is the object at a given height?
Architecture and construction: the cable of a suspension bridge and the sweep of a parabolic arch follow quadratics, and finding where the curve meets a support means solving one.
Business and economics: revenue and profit models are often quadratic, and the break-even points, where profit equals zero, are exactly the roots.
Computer graphics and games: collision detection and ray tracing solve quadratics to find where a moving line meets a circle or a sphere.
Optics and communication: parabolic dishes and reflectors focus light and signals, and their cross-section is a quadratic curve.
One formula finds landing points, break-even prices, and collision moments. Anywhere a curve bends back on itself, the Sridharacharya formula is close by.
What Are The Most Common Sridharacharya Formula Mistakes?
These three errors account for most lost marks, and they line up with the calculation mistakes documented in OpenStax's algebra text and in error studies of the quadratic formula.
Getting the sign of $b$ wrong.
Where it slips in:
When $b$ is negative, a student substitutes $-b$ as if it were still negative, forgetting that the formula already carries a minus sign in front of $b$.
Don't do this:
Do not copy $b$ straight into the $-b$ slot without flipping its sign.
The correct way:
Write $a$, $b$, $c$ with their signs first, then substitute. For $2x^2 - x - 3 = 0$, $b = -1$, so $-b = -(-1) = +1$, not $-1$.
Dividing only part of the numerator by $2a$.
Where it slips in:
After reaching $x = \frac{4 \pm 2\sqrt{3}}{2}$, a student divides only the first term and writes $2 \pm 2\sqrt{3}$, cancelling the $2$ with just one piece of the numerator.
Don't do this:
Do not simplify one term and leave the other untouched.
The correct way:
The whole numerator, both $-b$ and the root part, is divided by $2a$. So $\frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$, dividing every term.
Mishandling a negative discriminant.
Where it slips in:
When $b^2 - 4ac$ comes out negative, a student either takes the square root as if the number were positive, or stops and writes "no solution" without naming the complex roots.
Don't do this:
Do not take a real square root of a negative number, and do not read "no real roots" as "no roots at all."
The correct way:
Check the discriminant first. If $D < 0$, state that there are no real roots and two complex conjugate roots, and only take the square root of a non-negative number.
Practice Problems On The Sridharacharya Formula
Solve each with the formula, checking the discriminant before you substitute. Answers follow each line.
Solve $x^2 - 4x - 5 = 0$.
(Answer: $D = 36$, so $x = 5$ or $x = -1$.)Solve $x^2 + 4x + 3 = 0$.
(Answer: $D = 4$, so $x = -1$ or $x = -3$.)Solve $3x^2 - 5x + 2 = 0$.
(Answer: $D = 1$, so $x = 1$ or $x = \frac{2}{3}$.)Solve $2x^2 + 5x - 3 = 0$.
(Answer: $D = 49$, so $x = \frac{1}{2}$ or $x = -3$.)Solve $x^2 - 2x - 1 = 0$.
(Answer: $D = 8$, so $x = 1 + \sqrt{2}$ or $x = 1 - \sqrt{2}$.)Without solving, describe the roots of $x^2 + 2x + 5 = 0$.
(Answer: $D = 4 - 20 = -16 < 0$, so no real roots; two complex conjugate roots.)
Where Should You Go Next After The Sridharacharya Formula?
The Sridharacharya formula opens onto the wider world of quadratics, and a few natural doors lead out from here.
Quadratic equations. The full family of $ax^2 + bx + c = 0$, the different methods for solving them, and when to reach for each.
The discriminant. Go deeper on $b^2 - 4ac$ and learn to read the roots of any quadratic before solving it.
Completing the square. The method underneath the formula, useful on its own for vertex form and for graphing parabolas.
If your child is meeting quadratics for the first time, a live Bhanzu trainer builds the formula up from completing the square, so it is understood rather than memorised, in the Bhanzu algebra program.
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