An equation that can be written as $ax^2+bx+c=0$ with $a\ne0$ is quadratic. Learn to recognise one, check whether a number is a root, and turn everyday problems about areas, numbers and speeds into quadratic equations.
A school in Jaipur wants a rectangular vegetable garden. The length must be $3$ m more than twice the breadth, and the garden should cover exactly $90$ m². How wide should it be? Call the breadth $x$ metres. Then the length is $2x+3$ metres and the area condition says $x(2x+3)=90$, that is $2x^2+3x-90=0$. The unknown is multiplied by itself, so this is not a linear equation, and the methods of Chapter 3 cannot solve it.
Equations like this are called quadratic equations, and they appear whenever an area, a product of two unknown amounts or a speed-and-time relation is involved. In this section we learn three skills: recognising a quadratic equation (sometimes it is in disguise), checking whether a number is a root, and modelling a word problem by a quadratic equation in standard form. Solving them comes in Section 4.2 and Section 4.3.
What is a quadratic equation?
In Section 2.1 you met quadratic polynomials such as $2x^2+3x-90$. If we set a quadratic polynomial equal to zero, we get a quadratic equation.
📘 Definition 4.1 — Quadratic equation; standard form
A quadratic equation in the variable $x$ is an equation that can be written in the form
$$ax^2+bx+c=0,$$
where $a$, $b$, $c$ are real numbers and $a\ne0$. Written this way — all terms on the left in descending powers of $x$, and $0$ on the right — the equation is in standard form. In other words, a quadratic equation is $p(x)=0$ where $p(x)$ is a quadratic polynomial.
The condition $a\ne0$ is essential: if $a=0$ the $x^2$ term disappears and the equation is linear. The numbers $b$ and $c$ may be zero, so $x^2-9=0$ ($b=0$) and $5x^2+2x=0$ ($c=0$) are quadratic equations too.
⚠️ Simplify before you decide
An equation can contain $x^2$ (or even $x^3$) and still not be quadratic, because the high powers may cancel. And an equation can look harmless and turn out to be quadratic. Always expand, move everything to one side and collect like terms. Only then look at the highest power of $x$ that is left with a non-zero coefficient: if it is $2$, the equation is quadratic.
✏️ Example 1 — Is it quadratic?
Check whether each equation is a quadratic equation. If it is, write it in standard form.
(a) $(x-2)^2+1=2x-3$ (b) $x(x+1)+8=(x+2)(x-2)$ (c) $(x+2)^3=x^3-4$ (d) $(2x-1)(x-3)=(x+5)(x-1)$
💡 Show solution▶
(a) Left side: $x^2-4x+4+1=x^2-4x+5$. So $x^2-4x+5=2x-3$, which gives $x^2-6x+8=0$. Highest power $2$ with coefficient $1\ne0$: quadratic.
(b) Left side: $x^2+x+8$. Right side: $x^2-4$. So $x^2+x+8=x^2-4$, and the $x^2$ terms cancel: $x+12=0$. This is linear, not quadratic.
(c) $(x+2)^3=x^3+6x^2+12x+8$. So $x^3+6x^2+12x+8=x^3-4$; the $x^3$ terms cancel, leaving $6x^2+12x+12=0$, or (dividing by $6$) $x^2+2x+2=0$. Quadratic, even though it started with cubes.
A real number $\alpha$ is a root (or solution) of the quadratic equation $ax^2+bx+c=0$ if putting $x=\alpha$ makes the left side zero:
$$a\alpha^2+b\alpha+c=0 .$$
We say that $\alpha$ satisfies the equation. To solve the equation means to find all its roots. The roots of $ax^2+bx+c=0$ are exactly the zeroes of the polynomial $ax^2+bx+c$.
Because roots are zeroes, everything you learnt in Chapter 2 carries over. A zero of $ax^2+bx+c$ is the $x$-coordinate of a point where the parabola $y=ax^2+bx+c$ meets the $x$-axis, and a parabola meets a line at most twice.
📐 Theorem 4.1 — Roots and the graph
Let $a\ne0$. The real roots of $ax^2+bx+c=0$ are the $x$-coordinates of the points where the graph of $y=ax^2+bx+c$ meets the $x$-axis. Consequently a quadratic equation has at most two real roots: it may have two different roots, one root (the parabola just touches the axis), or no real root (the parabola stays above or below the axis).
✏️ Example 2 — Checking roots by substitution
Which of the numbers $2$, $-3$ and $-\tfrac52$ are roots of $2x^2+x-10=0$?
💡 Show solution▶
$x=2$ $2(2)^2+2-10=8+2-10=0$. So $2$ is a root.
$x=-3$ $2(-3)^2+(-3)-10=18-3-10=5\ne0$. So $-3$ is not a root.
$x=-\tfrac52$ $2\left(\tfrac{25}{4}\right)-\tfrac52-10=\tfrac{25}{2}-\tfrac52-10=10-10=0$. So $-\tfrac52$ is a root.
Check A quadratic equation has at most two roots, and we found two, so these are all of them. Also their sum $2-\tfrac52=-\tfrac12=-\tfrac ba$ and product $-5=\tfrac ca$, as in Section 2.2. ✓
✏️ Example 3 — An unknown coefficient
If $x=3$ is a root of $kx^2-5x-3=0$, find $k$. Then find the other root.
💡 Show solution▶
Substitute $x=3$ must make the left side zero: $k(3)^2-5(3)-3=0$, so $9k-18=0$ and $k=2$.
Other root The equation is $2x^2-5x-3=0$. The product of the roots is $\tfrac ca=-\tfrac32$. One root is $3$, so the other is $-\tfrac32\div3=-\tfrac12$.
Interactive 1 — Root Tester: Does This Number Satisfy the Equation?
Choose an equation and a trial value of $x$. The amber bar shows the value of the left side $ax^2+bx+c$ at that $x$: above the axis it is positive, below it is negative. A number is a root exactly when the bar shrinks to nothing and the point sits on the $x$-axis. Press Play to sweep $x$ from left to right; each root is marked in red as the sweep passes it.
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substitution—
left side at $x$—
roots marked—
Modelling situations with quadratic equations
A mathematical model of a situation is an equation (or set of equations) that captures its conditions. Building the model is often the hardest part of a word problem, so we practise it on its own here. The roots we check below were found by the methods of the next two sections.
🧭 Problem-solving strategy — from words to a quadratic equation
Read the problem twice. Decide what to call $x$, and write down its unit ("let the breadth be $x$ m").
Write every other quantity in terms of $x$ (length $2x+3$ m; next integer $x+1$; time $\tfrac{480}{x}$ h).
Find the sentence that gives a condition ("the area is $90$ m²", "it takes $3$ hours more") and turn it into an equation.
Expand, clear any fractions, and bring the equation to the standard form $ax^2+bx+c=0$. Divide by a common factor if there is one.
After solving, check each root against the situation: lengths, speeds and numbers of people must be positive, numbers of objects must be whole. Reject roots that make no sense.
✏️ Example 4 — The school garden
The length of a rectangular garden is $3$ m more than twice its breadth, and its area is $90$ m². Write a quadratic equation for the breadth. Check that $x=6$ is a root, and find the dimensions of the garden.
💡 Show solution▶
Unknown Let the breadth be $x$ m. Then the length is $(2x+3)$ m.
Equation Area: $x(2x+3)=90$, so $2x^2+3x=90$, i.e. $2x^2+3x-90=0$.
Check $x=6$: $2(36)+3(6)-90=72+18-90=0$. ✓ (The other root is $-\tfrac{15}{2}$, which is rejected: a breadth cannot be negative.)
Answer Breadth $6$ m and length $2(6)+3=15$ m. Check: $6\times15=90$ m². ✓
✏️ Example 5 — Consecutive integers
The product of two consecutive positive integers is $306$. Form a quadratic equation, check that $17$ is a root, and find the integers.
💡 Show solution▶
Unknown Let the smaller integer be $x$; the next one is $x+1$.
Equation $x(x+1)=306$, so $x^2+x-306=0$.
Check $17^2+17-306=289+17-306=0$. ✓ The equation has another root, $-18$ (check: $324-18-306=0$), but the integers must be positive, so we reject it.
Answer The integers are $17$ and $18$; indeed $17\times18=306$. ✓
✏️ Example 6 — A slower train
A train travels $480$ km at a uniform speed. If its speed had been $8$ km/h less, it would have taken $3$ hours more for the same journey. Write a quadratic equation for the speed, and check that $40$ km/h fits.
💡 Show solution▶
Unknown Let the speed be $x$ km/h. Time $=\dfrac{\text{distance}}{\text{speed}}$, so the actual time is $\dfrac{480}{x}$ h and the slower time is $\dfrac{480}{x-8}$ h.
Condition Slower time $-$ actual time $=3$: $\;\dfrac{480}{x-8}-\dfrac{480}{x}=3$.
Clear fractions Multiply by $x(x-8)$: $\;480x-480(x-8)=3x(x-8)$, so $3840=3x^2-24x$. Divide by $3$: $\;x^2-8x-1280=0$.
Check $40^2-8(40)-1280=1600-320-1280=0$. ✓ In words: at $40$ km/h the journey takes $\tfrac{480}{40}=12$ h; at $32$ km/h it takes $\tfrac{480}{32}=15$ h — exactly $3$ h more. (The other root, $-32$, is rejected because a speed here must be positive.)
Interactive 2 — Modelling a Park Border
A park is $20$ m long and $15$ m wide. A flower border of the same width $x$ m runs along all four sides inside the park, and the grass in the middle must have a chosen area $A$. The grass is $(20-2x)$ m by $(15-2x)$ m, so the model is $(20-2x)(15-2x)=A$, or $4x^2-70x+(300-A)=0$. Drag $x$ or press Play: on the right, the grass area falls as the border widens, and a root is a width where the curve meets the target line.
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grass (m)—
grass area (m²)—
standard form—
Exercises
I
Check whether each equation is a quadratic equation. If it is, write it in standard form.
1 $(x+1)^2=2(x-3)$
2 $x^2-2x=(-2)(3-x)$
3 $(x-2)(x+1)=(x-1)(x+3)$
4 $(x-3)(2x+1)=x(x+5)$
5 $x^2+3x+1=(x-2)^2$
6 $(x+2)^3=2x(x^2-1)$
7 $x^3-4x^2-x+1=(x-2)^3$
8 $3x^2-5x+7=3(x^2-2)$
💡 Show solutions▶
1 $x^2+2x+1=2x-6$ gives $x^2+7=0$. Quadratic ($a=1$, $b=0$, $c=7$). It happens to have no real root, since $x^2+7\ge7$ — but it is still a quadratic equation.
2 $x^2-2x=-6+2x$ gives $x^2-4x+6=0$. Quadratic.
3 $x^2-x-2=x^2+2x-3$ gives $-3x+1=0$. Not quadratic (linear).
4 $x=\tfrac12$: $\tfrac12+\tfrac52-3=0$, a root. $x=3$: $18+15-3=30\ne0$, not a root. (The other root is $-3$: $18-15-3=0$.)
5 $4+2k-10=0$, so $k=3$. Product of roots $=\tfrac{-10}{1}=-10$, so the other root is $-10\div2=-5$. Check: $25-15-10=0$ ✓.
6 $2\cdot\tfrac14-\tfrac k2-3=0$, so $\tfrac k2=-\tfrac52$ and $k=-5$. Product of roots $=-\tfrac32$, so the other root is $-\tfrac32\div\left(-\tfrac12\right)=3$. Check: $18-15-3=0$ ✓.
7 $12-14+p=0$, so $p=2$. Product of roots $=\tfrac23$, so the other root is $\tfrac23\div(-2)=-\tfrac13$. Check: $3\cdot\tfrac19-\tfrac73+2=\tfrac13-\tfrac73+2=0$ ✓.
8 $x=2$: $4a+2+b=0$. $x=-3$: $9a-3+b=0$. Subtracting the first from the second: $5a-5=0$, so $a=1$ and then $b=-6$. The equation is $x^2+x-6=0$; check: $4+2-6=0$ and $9-3-6=0$ ✓.
III
Modelling. Represent each situation by a quadratic equation in standard form. Then check that the given value is a root and answer the question.
1 A rectangular plot has area $528$ m², and its length is $1$ m more than twice its breadth. (Check breadth $16$ m.) Find its dimensions.
2 The product of two consecutive positive even integers is $288$. (Check $16$.) Find the integers.
3 The sum of a natural number and its square is $156$. (Check $12$.)
4 The product of Meena's age $3$ years ago and her age $4$ years from now is $170$. (Check $13$.) How old is she now?
5 The hypotenuse of a right triangle is $25$ cm and one leg is $17$ cm longer than the other. (Check $7$.) Find the legs.
6 A bus travels $360$ km. If its speed were $10$ km/h more, it would take $3$ hours less. (Check $30$ km/h.)
7 Some students share ₹ $1200$ equally for a picnic. If $5$ more students join, each pays ₹ $8$ less. (Check $25$ students.) How much does each pay at first?
8 A stone is thrown upwards from the top of a building. Its height above the ground after $t$ seconds is $h=35+30t-5t^2$ metres. When does it hit the ground? (Check $t=7$.)
💡 Show solutions▶
1 Breadth $x$ m, length $(2x+1)$ m: $x(2x+1)=528$, so $2x^2+x-528=0$. Check: $2(256)+16-528=0$ ✓. The plot is $16$ m by $33$ m ($16\times33=528$ ✓).
2 Integers $x$ and $x+2$: $x^2+2x-288=0$. Check: $256+32-288=0$ ✓. The integers are $16$ and $18$.
3 $x^2+x=156$, so $x^2+x-156=0$. Check: $144+12-156=0$ ✓. The number is $12$. (The other root, $-13$, is not a natural number.)
4 Present age $x$ years: $(x-3)(x+4)=170$, so $x^2+x-12=170$ and $x^2+x-182=0$. Check: $169+13-182=0$ ✓. Meena is $13$ years old ($10\times17=170$ ✓).
5 Legs $x$ and $x+17$ cm. Pythagoras: $x^2+(x+17)^2=625$, so $2x^2+34x-336=0$, i.e. $x^2+17x-168=0$. Check: $49+119-168=0$ ✓. Legs $7$ cm and $24$ cm ($49+576=625$ ✓).
6 Speed $x$ km/h: $\dfrac{360}{x}-\dfrac{360}{x+10}=3$. Multiply by $x(x+10)$: $3600=3x(x+10)$, so $x^2+10x-1200=0$. Check: $900+300-1200=0$ ✓. Speed $30$ km/h: $12$ h, versus $9$ h at $40$ km/h ✓.
7 $x$ students: $\dfrac{1200}{x}-\dfrac{1200}{x+5}=8$. Multiply by $x(x+5)$: $6000=8x(x+5)$, so $x^2+5x-750=0$. Check: $625+125-750=0$ ✓. At first each pays ₹ $\tfrac{1200}{25}=48$ (then ₹ $\tfrac{1200}{30}=40$, which is ₹ $8$ less ✓).
8 Ground means $h=0$: $35+30t-5t^2=0$, i.e. $5t^2-30t-35=0$, or $t^2-6t-7=0$. Check: $49-42-7=0$ ✓. The stone lands after $7$ s. (The other root, $t=-1$, is before the throw, so it is rejected.)
IV
Think deeper.
1 For which value of $k$ is $(k-2)x^2+5x+1=0$ not a quadratic equation?
2 Write a quadratic equation in standard form whose roots are $4$ and $-3$.
3 Write a quadratic equation with whole-number coefficients whose roots are $\tfrac12$ and $-2$.
4 Riya says she has found three different roots of a quadratic equation. Explain why she must have made a mistake.
5 Show that $x=1$ is a root of $ax^2+bx+c=0$ whenever $a+b+c=0$. Use this to find both roots of $7x^2-12x+5=0$.
6 Explain why $x=0$ is a root of $ax^2+bx+c=0$ exactly when $c=0$.
💡 Show solutions▶
1 When $k-2=0$, i.e. $k=2$, the equation becomes $5x+1=0$, which is linear.
2 Sum $=1$, product $=-12$: $\;x^2-x-12=0$. Check: $16-4-12=0$ and $9+3-12=0$ ✓.
3 Sum $=-\tfrac32$, product $=-1$: $\;x^2+\tfrac32x-1=0$; multiply by $2$: $\;2x^2+3x-2=0$. Check: $\tfrac12+\tfrac32-2=0$ and $8-6-2=0$ ✓.
4 By Theorem 4.1 a quadratic equation has at most two real roots (a parabola meets the $x$-axis at most twice), so three different roots are impossible.
5 At $x=1$ the left side is $a+b+c$, which is $0$. For $7x^2-12x+5$: $7-12+5=0$, so $x=1$ is a root. Product of roots $=\tfrac57$, so the other root is $\tfrac57$. Check: $7\cdot\tfrac{25}{49}-\tfrac{60}{7}+5=\tfrac{25}{7}-\tfrac{60}{7}+\tfrac{35}{7}=0$ ✓.
6 At $x=0$ the left side is $a\cdot0+b\cdot0+c=c$. So $0$ is a root if and only if $c=0$ (and then $ax^2+bx=x(ax+b)$).
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study
How this is examined: 1-mark questions ask you to spot the quadratic equation in a list or to find a coefficient from a given root; 2- and 3-mark questions ask you to "represent the situation mathematically". Always state what $x$ stands for (with its unit) and show the simplification to standard form — those lines earn the marks.
E1.Board · 1 mark Which of the following is a quadratic equation? (A) $x^2+3x=(x-1)^2$ (B) $(x+1)^3=x^3+2$ (C) $x(x-4)=x^2+7$ (D) $2x+3=x(x+1)-x^2$
Trap (A) and (C) contain $x^2$ on both sides, which tempts you to say "quadratic" — but the $x^2$ terms cancel.
Answer (B) — the cubes cancel and an $x^2$ term with coefficient $3\ne0$ remains.
E2.Board · 1 mark If $x=-\tfrac12$ is a root of $3x^2+2kx-3=0$, then $k$ equals (A) $\tfrac94$ (B) $-\tfrac94$ (C) $\tfrac34$ (D) $-\tfrac34$
💡 Show solution▶
Substitute $3\cdot\tfrac14+2k\left(-\tfrac12\right)-3=0$, so $\tfrac34-k-3=0$ and $k=-\tfrac94$.
Trap (A) comes from losing the minus sign in $2k\left(-\tfrac12\right)=-k$.
Answer (B) $-\tfrac94$. Check: $3x^2-\tfrac92x-3=0$, i.e. $2x^2-3x-2=0$, and $2\cdot\tfrac14+\tfrac32-2=0$ ✓.
E3.Board · 1 markAssertion (A): $(x-2)^2-3=x^2+1$ is not a quadratic equation. Reason (R): An equation is quadratic only if, after it is simplified to the form $ax^2+bx+c=0$, the coefficient $a$ is not zero. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
💡 Show solution▶
Check A $x^2-4x+4-3=x^2+1$ gives $-4x=0$: linear. A is true.
Check R R is Definition 4.1, so it is true, and it is exactly the reason the equation fails to be quadratic.
Answer (A). Choosing (D) means judging by the look of the equation (it has $x^2$) instead of simplifying.
E4.Board · 2 marks The sum of the squares of two consecutive odd natural numbers is $394$. Represent this by a quadratic equation, and verify that the smaller number is $13$.
💡 Show solution▶
Model Let the numbers be $x$ and $x+2$. Then $x^2+(x+2)^2=394$, so $2x^2+4x+4=394$, i.e. $x^2+2x-195=0$. (1 mark)
Verify $13^2+2(13)-195=169+26-195=0$ ✓, so the numbers are $13$ and $15$; indeed $169+225=394$. (1 mark)
E5.Board · 3 marks A cyclist covers $36$ km at a uniform speed. Had she cycled $3$ km/h faster, she would have taken $1$ hour less. Show that her speed $x$ (in km/h) satisfies $x^2+3x-108=0$, and verify that her speed is $9$ km/h.
💡 Show solution▶
Times Actual time $\dfrac{36}{x}$ h, faster time $\dfrac{36}{x+3}$ h, and $\dfrac{36}{x}-\dfrac{36}{x+3}=1$. (1 mark)
Simplify Multiply by $x(x+3)$: $36(x+3)-36x=x(x+3)$, so $108=x^2+3x$, i.e. $x^2+3x-108=0$. (1 mark)
Verify $81+27-108=0$ ✓. At $9$ km/h the ride takes $4$ h; at $12$ km/h it takes $3$ h — one hour less ✓. (The other root, $-12$, is rejected.) (1 mark)
E6.Board · Case studyThe society park. A housing society in Chennai has a rectangular park $30$ m long and $20$ m wide. A walking track of uniform width $x$ m is to be laid inside the park along all four sides, leaving a lawn of area $336$ m² in the middle. (i) Write the length and breadth of the lawn in terms of $x$. (1 mark) (ii) Show that $x$ satisfies $x^2-25x+66=0$. (1 mark) (iii) Verify that both $x=3$ and $x=22$ satisfy this equation, and decide which one is the width of the track, giving a reason. (2 marks)
💡 Show solution▶
(i) The track takes $x$ m from each end: lawn length $(30-2x)$ m, lawn breadth $(20-2x)$ m.
(ii) $(30-2x)(20-2x)=336\Rightarrow600-100x+4x^2=336\Rightarrow4x^2-100x+264=0$. Dividing by $4$: $\;x^2-25x+66=0$.
(iii) Decide With $x=22$ the lawn breadth would be $20-44=-24$ m, which is impossible, so $x=22$ is rejected. The track is $3$ m wide (lawn $24$ m × $14$ m $=336$ m² ✓).
🔬 Research–Teaching NexusMini project · optional, about 45 minutes
Research question. Does a falling ball really obey a quadratic model? If a ball is dropped from a height of $H$ metres, physicists model the time $t$ (in seconds) it takes to reach the ground by the quadratic equation $4.9t^2-H=0$. Do your own measurements give roots of this equation?
Why it matters. Engineers who design drop tests for phones, sports scientists who study the flight of a cricket or tennis ball, and space agencies such as ISRO that plan the descent of a capsule all start from quadratic models like this one. A model is only useful if measurements agree with it — checking that agreement is what scientists do every day.
What to do.
With an adult's help, mark heights of $1$ m, $2$ m and (from a first-floor balcony or staircase, safely) about $3$ m. Measure each height with a tape measure.
Drop a small ball from each height and film it with a phone's slow-motion camera (many phones record $240$ frames per second). Count the frames from release to impact and divide by the frame rate to get $t$. Repeat three times for each height and take the average.
For each height write the model equation $4.9t^2-H=0$ in standard form, e.g. $4.9t^2-2=0$ for $H=2$.
Substitute your measured $t$. How close to $0$ is the left side? Express the difference as a percentage of $H$.
Find, by trial in a spreadsheet, the positive value of $t$ that makes the left side as close to $0$ as possible, and compare it with your measurement. Why is the negative root of no use here?
Tools: Tape measure, a small ball, a phone with slow-motion video, a spreadsheet or calculator
Hand in: A table of $H$, the three measured times, their average, the value of $4.9t^2-H$ and the predicted root; plus four or five sentences on how well the quadratic model fits and what might cause the differences (reaction time, air resistance, counting frames).
Going further. Drop a crumpled paper ball and a flat sheet of paper from the same height. Does the quadratic model fit both equally well? What does that tell you about when a model can be trusted?
Teacher/parent note: the positive roots of $4.9t^2-H=0$ are about $0.45$ s for $H=1$ m, $0.64$ s for $H=2$ m and $0.78$ s for $H=3$ m. Students only need substitution and trial here; the equation can be solved exactly once square roots or the quadratic formula (Section 4.3) are used.
🧪 Self-check quiz
0/5 correct
Q1. Which of these is a quadratic equation?
Q2. The standard form of $(x+3)(x-2)=4x$ is
Q3. Which number is a root of $x^2-7x+12=0$?
Q4. If $x=2$ is a root of $x^2+kx+6=0$, then $k$ is
Q5. A rectangle's length is $4$ cm more than its breadth $x$ cm, and its area is $96$ cm². Which equation models this?