Quadratic Equations

The chapter builds in four steady stages, and each stage leans on the one before it. It opens by defining a quadratic equation precisely: the standard form $ax^2+bx+c=0$, $a\neq0$, and the care needed to simplify an equation before deciding whether it truly is quadratic (an equation can look quadratic and cancel down to linear, or look cubic and collapse into quadratic — NCERT's own worked Example 2 shows both traps). It then moves to solving by factorisation, the technique of splitting the middle term $bx$ into two parts that multiply, after regrouping, into two linear factors — practised on rising difficulty, including a case with a repeated root. Because factorisation depends on spotting a lucky split, the chapter next derives the quadratic formula by completing the square on the fully general equation, giving a method that solves any quadratic equation, factorable or not. The chapter closes with the discriminant and the nature of roots: reading the sign of $b^2-4ac$ to know, without solving, how many real answers an equation has — and applying that test to a geometric problem (a pole between two gates on a circular park) where the discriminant decides whether the construction is even possible.

Recognising a quadratic equation: the standard form

A quadratic equation in the variable $x$ is an equation that can be written as

where: $a$ is the coefficient of $x^2$, $b$ is the coefficient of $x$, and $c$ is the constant term. The condition $a \neq 0$ is essential — if $a=0$ the $x^2$ term vanishes and the equation collapses to the linear equation $bx+c=0$, which is a completely different object with at most one root. More generally, any equation of the form $p(x) = 0$ where $p(x)$ is a polynomial of degree $2$ is a quadratic equation; writing its terms in descending powers of $x$ gives the standard form shown above. The prayer-hall equation $2x^2+x-300=0$ is already in standard form with $a=2$, $b=1$, $c=-300$. The catch, which NCERT flags explicitly, is that an equation must often be simplified first before you can tell whether it is quadratic at all. Two examples make the point in opposite directions. The equation $x(x+1)+8=(x+2)(x-2)$ expands to $x^2+x+8 = x^2-4$, and the $x^2$ terms cancel on both sides, leaving $x+12=0$ — a linear equation dressed up to look quadratic. On the other hand, $(x+2)^3 = x^3-4$ expands to $x^3+6x^2+12x+8=x^3-4$, and the $x^3$ terms cancel, leaving $6x^2+12x+12=0$, i.e. $x^2+2x+2=0$ — a genuinely quadratic equation hiding inside what looked like a cubic. The lesson: always expand and collect terms fully before classifying an equation by its degree.

Roots, zeroes, and how many answers to expect

A real number $\alpha$ is called a root of $ax^2+bx+c=0$ if it satisfies the equation, that is, if $a\alpha^2+b\alpha+c=0$. Saying "$\alpha$ is a root of the equation" and "$\alpha$ is a zero of the quadratic polynomial $ax^2+bx+c$" describe exactly the same fact, connecting this chapter directly to the polynomials chapter. Since a quadratic polynomial can have at most two zeroes, a quadratic equation can have at most two roots — never three, never infinitely many. Checking that a given number is a root is pure substitution: for $2x^2-3x+1=0$, putting $x=1$ gives $2(1)-3(1)+1=0$, so $x=1$ is a root.

Method one: solving by factorisation

When $ax^2+bx+c$ can be written as a product of two linear factors, the equation is solved instantly by the zero-product rule: if a product of two real numbers is zero, at least one of them must be zero. The standard technique is splitting the middle term — writing $b$ as a sum $p+q$ chosen so that $p \cdot q = ac$, then regrouping the four terms into two pairs that share a common factor. For $2x^2-5x+3=0$, the middle term $-5x$ splits as $-2x-3x$ because $(-2)\times(-3)=6=2\times3=ac$; grouping gives $2x(x-1)-3(x-1) = (2x-3)(x-1)$, so the roots are $x=\tfrac32$ and $x=1$. The same idea handles a repeated root: for $3x^2-2\sqrt6\,x+2=0$, the left side is exactly the perfect square $(\sqrt3\,x-\sqrt2)^2$, so the equation becomes $(\sqrt3\,x-\sqrt2)^2=0$, forcing $x=\dfrac{\sqrt2}{\sqrt3}=\dfrac{\sqrt6}{3}$ — the same value twice. Factorisation is fast when it works, but it depends on spotting the right split, and many quadratics with irrational or awkward coefficients resist it entirely.

Method two: the quadratic formula, for every quadratic equation

Because factorisation is not guaranteed to work, mathematicians sought a method that solves any quadratic equation using only its coefficients. Completing the square on the fully general equation $ax^2+bx+c=0$ (worked out in full below) produces the single most quoted result in school algebra:

where: $a$, $b$, $c$ are the coefficients of the equation in standard form, and the $\pm$ generates two roots at once — one using $+$, one using $-$ — from a single formula. This is credited to Sridharacharya, who derived it by completing the square. Its power is that it needs no cleverness: plug in $a$, $b$, $c$ and simplify. Applied to $2x^2-5x+3=0$: $x = \dfrac{5 \pm \sqrt{25-24}}{4} = \dfrac{5\pm1}{4}$, giving $x=\tfrac32$ or $x=1$ — exactly the roots found by factorisation above, confirming the two methods agree.

The discriminant: the nature of roots, before you solve

Look again at the formula: everything under the square root sign, $b^2-4ac$, controls whether a real square root even exists. This quantity is important enough to have its own name.

where: $D$ is the discriminant of $ax^2+bx+c=0$. Because a real number's square is never negative, $\sqrt{D}$ is a real number exactly when $D \geq 0$; this single sign check tells you, before any further arithmetic, how many real roots the equation has. If $D>0$, the formula gives two genuinely different values, $\dfrac{-b+\sqrt D}{2a}$ and $\dfrac{-b-\sqrt D}{2a}$ — two distinct real roots. If $D=0$, the $\pm\sqrt D$ term vanishes and both signs collapse to the same value $-\dfrac{b}{2a}$ — two equal (repeated) real roots. If $D<0$, no real number squares to a negative number, so $\sqrt D$ does not exist as a real number — the equation has no real roots. Graphically, $y=ax^2+bx+c$ is a parabola, and its roots are exactly where that parabola crosses the $x$-axis; the discriminant's sign is simply telling you how many times a parabola meets a straight line.

Board level - the prayer hall dimensions (NCERT Example 6, factorisation)

The breadth of a rectangular prayer hall is $x$ metres and its length is $(2x+1)$ metres. If the carpet area is $300\ \text{m}^2$, find the length and breadth of the hall.

Step 1. Model the area condition: $(2x+1)\,x = 300$, which expands to $2x^2+x=300$, i.e. $2x^2+x-300=0$ in standard form, with $a=2,\,b=1,\,c=-300$.

Step 2. Solve by splitting the middle term. We need two numbers whose sum is $1$ (the coefficient of $x$) and whose product is $ac = 2\times(-300)=-600$. The pair $-24$ and $25$ works, since $-24+25=1$ and $(-24)(25)=-600$: $2x^2-24x+25x-300=0$.

Step 3. Group and factor: $2x(x-12)+25(x-12)=0$, so $(x-12)(2x+25)=0$, giving $x=12$ or $x=-\dfrac{25}{2}=-12.5$.

Step 4. Reject the physically impossible root. Since $x$ is a breadth, it cannot be negative, so $x=-12.5$ is discarded. The breadth is $x=12$ m, and the length is $2(12)+1=25$ m. Answer: breadth $12$ m, length $25$ m.

Board level - the pole between two gates (NCERT Example 8, discriminant and quadratic formula)

A pole is to be erected on the boundary of a circular park of diameter $13$ m, so that the difference of its distances from two diametrically opposite gates $A$ and $B$ is $7$ m. Is this possible, and if so, at what distances from the two gates should the pole stand?

Step 1. Let $BP=x$ m be the distance of the pole $P$ from gate $B$; then $AP=(x+7)$ m. Since $AB$ is a diameter and $P$ lies on the circle, the angle $\angle APB = 90°$ (angle in a semicircle), so by the Pythagoras theorem $AP^2+PB^2=AB^2$: $(x+7)^2+x^2 = 13^2$.

Step 2. Expand and simplify: $x^2+14x+49+x^2=169$, i.e. $2x^2+14x-120=0$, which divides through by $2$ to give the standard form $x^2+7x-60=0$, with $a=1,\,b=7,\,c=-60$.

Step 3. Check the discriminant before solving, since the question asks whether it is even possible: $D=b^2-4ac = 7^2-4(1)(-60)=49+240=289>0$. Since $D>0$, two distinct real roots exist, so erecting the pole is indeed possible.

Step 4. Apply the quadratic formula: $x = \dfrac{-7\pm\sqrt{289}}{2} = \dfrac{-7\pm17}{2}$, giving $x=5$ or $x=-12$. Since $x$ is a distance, it must be positive, so $x=-12$ is rejected. Answer: the pole stands $5$ m from gate $B$ and $12$ m from gate $A$.

Board HOTS / foundation JEE-NEET - the train's speed (forming and solving a rate quadratic)

A train travels a distance of $480$ km at a uniform speed. If the speed had been $8$ km/h less, it would have taken $3$ hours more to cover the same distance. Find the speed of the train.

Step 1. Let the original speed be $x$ km/h ($x>8$, so the reduced speed is positive). Time taken at speed $x$ is $\dfrac{480}{x}$ hours; time taken at speed $(x-8)$ is $\dfrac{480}{x-8}$ hours, and the second time exceeds the first by $3$ hours: $\dfrac{480}{x-8} - \dfrac{480}{x} = 3$.

Step 2. Combine the fractions over a common denominator $x(x-8)$: $\dfrac{480x - 480(x-8)}{x(x-8)} = 3$, and the numerator simplifies since the $480x$ terms cancel: $480x - 480x + 3840 = 3840$. So $\dfrac{3840}{x(x-8)}=3$, which gives $3840 = 3x(x-8) = 3x^2-24x$.

Step 3. Divide by $3$ and rearrange into standard form: $x^2-8x-1280=0$, with $a=1,\,b=-8,\,c=-1280$. The discriminant is $D=(-8)^2-4(1)(-1280)=64+5120=5184=72^2$, a perfect square, so the formula gives rational roots: $x=\dfrac{8\pm72}{2}$, i.e. $x=40$ or $x=-32$.

Step 4. Reject the negative root, since a speed cannot be negative: $x=-32$ is discarded, leaving $x=40$. Verify: at $40$ km/h the journey takes $480/40=12$ h; at $32$ km/h it takes $480/32=15$ h, a difference of exactly $3$ h. Answer: the train's original speed is $40$ km/h. Timing note: aim for under four minutes. The trap is entirely in Step 1 — writing the slower speed's time as the larger fraction, $480/(x-8)$, not $480/x$; getting this the wrong way round flips the sign of the final linear equation and produces a wrong quadratic. Once the equation is set up correctly, recognising $5184=72^2$ quickly (or simply computing $\sqrt{5184}$ by trial, since it must be close to $70^2=4900$ and $75^2=5625$) saves time over long division.

• Not simplifying before judging the degree. An equation can look quadratic and reduce to linear (terms in $x^2$ cancel), or look cubic and reduce to quadratic (terms in $x^3$ cancel) — always expand and collect fully before deciding.
• Forgetting $a \neq 0$. If the coefficient of $x^2$ turns out to be zero after simplification, the equation is linear, not quadratic, and has at most one root, not two.
• Splitting on the wrong product. The two numbers used to split the middle term $bx$ must multiply to $ac$, not to $c$ alone and not to $b$ alone — this is the single most common factorisation error.
• Sign slips in the quadratic formula. Writing $x=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}$ instead of $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, especially when $b$ itself is negative, silently flips both roots' signs.
• Treating "no real roots" as a mistake. When $D<0$, "no real roots" is a complete and correct final answer — it does not mean the working went wrong; it means the parabola never crosses the $x$-axis.
• Confusing equal roots with two separate answers that happen to match. When $D=0$, the equation genuinely has one repeated root, conventionally still written as two equal values (e.g. $\tfrac13,\tfrac13$), not a single unrelated coincidence.
• Accepting both roots blindly in a word problem. After solving, check every root against the physical meaning of $x$ — lengths, ages, speeds, and counts of objects cannot be negative, and the inadmissible root must be explicitly discarded, not silently ignored.
• Leaving surds unrationalised. Roots like $\dfrac{\sqrt2}{\sqrt3}$ should be simplified to $\dfrac{\sqrt6}{3}$ before being called a final answer.

• Write any quadratic equation in the standard form $ax^2+bx+c=0$ and identify $a$, $b$, $c$ correctly, after fully simplifying both sides first.
• Decide, by expanding and simplifying, whether a given equation is genuinely quadratic, linear, or of some other degree.
• Translate a word problem — areas, ages, consecutive integers, speeds — into a quadratic equation in one variable.
• Solve a quadratic equation by splitting the middle term, choosing the split so the two numbers multiply to $ac$ and add to $b$.
• Derive and apply the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, including for equations that resist factorisation.
• Compute the discriminant $b^2-4ac$ and state, without solving further, whether the equation has two distinct real roots, two equal real roots, or no real roots.
• Recognise and correctly report a repeated (equal) root, including when the coefficients involve surds.
• In an applied problem, check each algebraic root against the real-world constraints on the variable and reject any inadmissible root with a clear reason.

Ask the tutor: if you are stuck deciding whether an equation is really quadratic, or which root to keep in a word problem, ask the tutor to walk through your specific equation step by step.