A qua­dratic equa­tion is an equa­tion in which the high­est power of the unknown is 2, such as 2x2+x−300=02x^2 + x - 300 = 0. Qua­dratic equa­tions turn up when­ever two unknown quan­ti­ties are mul­ti­plied together: the area of a room, the prod­uct of two ages, the total cost when the price depends on the quan­tity, or the time taken when speed changes. This les­son, based on Chap­ter 4 of the NCERT Class 10 text­book, explains what makes an equa­tion qua­dratic, how to write it in stan­dard form, how to turn a word prob­lem into one, and how to test whether an untidy equa­tion is really qua­dratic.

A prob­lem that leads to a qua­dratic equa­tion

The prayer hall

A char­ity trust wants to build a prayer hall. The car­pet area must be exactly 300 square metres, and the length of the hall must be one metre more than twice its breadth. What should the length and breadth be?

Give the unknown a let­ter. Let the breadth of the hall be xx metres. The length is "one metre more than twice the breadth", so the length is (2x+1)(2x + 1) metres. Now use the area:

area=length×breadth=(2x+1)×x=2x2+x\text{area} = \text{length} \times \text{breadth} = (2x+1) \times x = 2x^2 + x

The area is 300 square metres, so 2x2+x=3002x^2 + x = 300. Mov­ing every term to one side gives

2x2+x−300=02x^2 + x - 300 = 0

Rectangle drawn to scale, breadth x m (12 m) and length 2x + 1 m (25 m), labelled Area = 300 square metres and x(2x + 1) = 300.
The prayer hall: breadth xx m, length (2x+1)(2x+1) m, area 300 m². It is drawn at the true answer, x=12x = 12, which the next les­son finds.

This equa­tion is dif­fer­ent from the lin­ear equa­tions you already know, such as 3x+5=203x + 5 = 20, where xx appears only to the first power. Here xx is squared, and that changes how the equa­tion must be solved. That is why such equa­tions get their own name and their own chap­ter.

Def­i­n­i­tion and stan­dard form

What a qua­dratic equa­tion is

A qua­dratic equa­tion in the vari­able xx is an equa­tion of the form

ax2+bx+c=0,a≠0ax^2 + bx + c = 0, \qquad a \neq 0

where aa, bb and cc are real num­bers. The con­di­tion a≠0a \neq 0 is essen­tial. If aa were 0, the x2x^2 term would dis­ap­pear and we would be left with bx+c=0bx + c = 0, which is a lin­ear equa­tion.

More gen­er­ally, any equa­tion of the form p(x)=0p(x) = 0, where p(x)p(x) is a poly­no­mial of degree 2, is a qua­dratic equa­tion.

The names of the parts

  • aa is the coef­fi­cient of x2x^2 (also called the lead­ing coef­fi­cient);
  • bb is the coef­fi­cient of xx;
  • cc is the con­stant term.

When the terms are writ­ten in descend­ing order of pow­ers and set equal to zero, the equa­tion is in stan­dard form. The prayer hall equa­tion 2x2+x−300=02x^2 + x - 300 = 0 is already in stan­dard form, with a=2a = 2, b=1b = 1 and c=−300c = -300.

Recog­nis­ing the shape

The equa­tions 2x2−3x+1=02x^2 - 3x + 1 = 0 and 4x−3x2+2=04x - 3x^2 + 2 = 0 are both qua­dratic. The sec­ond is not writ­ten in stan­dard form, because the squared term is not first. Rear­rang­ing from the high­est power to the low­est gives −3x2+4x+2=0-3x^2 + 4x + 2 = 0, so a=−3a = -3, b=4b = 4 and c=2c = 2. If you pre­fer a pos­i­tive lead­ing coef­fi­cient, mul­ti­ply every term by −1-1 to get 3x2−4x−2=03x^2 - 4x - 2 = 0; this is the same equa­tion, with the same roots.

Why "at most two" mat­ters

A qua­dratic poly­no­mial has at most two zeroes, so a qua­dratic equa­tion has at most two roots. On a graph, the curve y=ax2+bx+cy = ax^2 + bx + c (a parabola) meets the xx-axis at most twice. You will use this idea again when you study the dis­crim­i­nant.

Graph of the parabola y = x squared minus 6x plus 8 on a grid, crossing the x-axis at the points (2, 0) and (4, 0), with its lowest point (3, -1) marked.
The graph of y=x2−6x+8y = x^2 - 6x + 8 crosses the xx-axis at x=2x = 2 and x=4x = 4: two roots, and no more.

How to test whether an equa­tion is qua­dratic

The method, step by step

  1. Expand every bracket on both sides, using iden­ti­ties such as (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2 and (a+b)3=a3+3a2b+3ab2+b3(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.
  2. Bring all terms to one side so that the other side is 0.
  3. Col­lect like terms and sim­plify.
  4. Look at the high­est power of xx that sur­vives. If it is x2x^2 with a non-zero coef­fi­cient, the equa­tion is qua­dratic. If the x2x^2 terms can­cel and only xx remains, it is lin­ear. If an x3x^3 term sur­vives, it is cubic.

Four worked checks

Exam­ple 1. Is (x−2)2+1=2x−3(x-2)^2 + 1 = 2x - 3 a qua­dratic equa­tion?

Expand the left-hand side: (x−2)2+1=x2−4x+4+1=x2−4x+5(x-2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5. So the equa­tion is x2−4x+5=2x−3x^2 - 4x + 5 = 2x - 3. Bring­ing every­thing to the left:

x2−4x−2x+5+3=0  ⟹  x2−6x+8=0x^2 - 4x - 2x + 5 + 3 = 0 \implies x^2 - 6x + 8 = 0

This has the form ax2+bx+c=0ax^2 + bx + c = 0 with a=1≠0a = 1 \neq 0. It is a qua­dratic equa­tion. (Its graph is the one shown above.)

Exam­ple 2. Is x(x+1)+8=(x+2)(x−2)x(x+1) + 8 = (x+2)(x-2) a qua­dratic equa­tion?

The left-hand side is x2+x+8x^2 + x + 8 and the right-hand side is x2−4x^2 - 4. So x2+x+8=x2−4x^2 + x + 8 = x^2 - 4. The x2x^2 term appears on both sides with the same coef­fi­cient, so it can­cels:

x+8+4=0  ⟹  x+12=0x + 8 + 4 = 0 \implies x + 12 = 0

No x2x^2 term is left. This is a lin­ear equa­tion. It is not a qua­dratic equa­tion, even though it looked like one at first.

Exam­ple 3. Is x(2x+3)=x2+1x(2x+3) = x^2 + 1 a qua­dratic equa­tion?

The left-hand side is 2x2+3x2x^2 + 3x, so 2x2+3x=x2+12x^2 + 3x = x^2 + 1, which rearranges to

2x2−x2+3x−1=0  ⟹  x2+3x−1=02x^2 - x^2 + 3x - 1 = 0 \implies x^2 + 3x - 1 = 0

Here 2x22x^2 and x2x^2 do not can­cel com­pletely. It is a qua­dratic equa­tion.

Exam­ple 4. Is (x+2)3=x3−4(x+2)^3 = x^3 - 4 a qua­dratic equa­tion?

Using the cube iden­tity, (x+2)3=x3+3x2(2)+3x(22)+23=x3+6x2+12x+8(x+2)^3 = x^3 + 3x^2(2) + 3x(2^2) + 2^3 = x^3 + 6x^2 + 12x + 8. So

x3+6x2+12x+8=x3−4x^3 + 6x^2 + 12x + 8 = x^3 - 4

The x3x^3 terms can­cel, leav­ing 6x2+12x+12=06x^2 + 12x + 12 = 0. Divid­ing every term by 6:

x2+2x+2=0x^2 + 2x + 2 = 0

The equa­tion looked cubic, but after sim­pli­fi­ca­tion it is a qua­dratic equa­tion. The les­son from all four checks: never judge an equa­tion by its appear­ance. Expand, sim­plify, and only then decide.

Turn­ing a word prob­lem into a qua­dratic equa­tion

The method, step by step

  1. Choose a let­ter for one unknown quan­tity and say clearly what it stands for, with units.
  2. Write every other unknown quan­tity in terms of that let­ter.
  3. Find the con­di­tion in the prob­lem that has not yet been used, and write it as an equa­tion.
  4. Expand, bring all terms to one side and write the result in stan­dard form.

Worked exam­ple: mar­bles

Exam­ple 5. John and Jivanti together have 45 mar­bles. Each of them loses 5 mar­bles, and the prod­uct of the num­bers they now have is 124. Form an equa­tion to find how many mar­bles each had at the start.

Let John have had xx mar­bles. Then Jivanti had 45−x45 - x. After los­ing 5 each, John has x−5x - 5 and Jivanti has 45−x−5=40−x45 - x - 5 = 40 - x. The prod­uct is 124:

(x−5)(40−x)=124(x-5)(40-x) = 124

Expand the left-hand side care­fully:

(x−5)(40−x)=40x−x2−200+5x=−x2+45x−200(x-5)(40-x) = 40x - x^2 - 200 + 5x = -x^2 + 45x - 200

So −x2+45x−200=124-x^2 + 45x - 200 = 124, that is, −x2+45x−324=0-x^2 + 45x - 324 = 0. Mul­ti­ply­ing every term by −1-1 to make the lead­ing coef­fi­cient pos­i­tive:

x2−45x+324=0x^2 - 45x + 324 = 0

This is the required qua­dratic equa­tion. Solv­ing it in the next les­son gives x=9x = 9 or x=36x = 36.

Worked exam­ple: toys

Exam­ple 6. A cot­tage indus­try makes a cer­tain num­ber of toys in a day. The cost of pro­duc­ing each toy (in rupees) is 55 minus the num­ber of toys made that day. On a par­tic­u­lar day the total cost of pro­duc­tion was ₹750. Form an equa­tion to find the num­ber of toys made that day.

Let xx be the num­ber of toys made that day. The cost of each toy is ₹(55−x)(55 - x), so the total cost is x(55−x)x(55 - x) rupees. There­fore

x(55−x)=75055x−x2=750−x2+55x−750=0x2−55x+750=0\begin{aligned} x(55-x) &= 750 \\ 55x - x^2 &= 750 \\ -x^2 + 55x - 750 &= 0 \\ x^2 - 55x + 750 &= 0 \end{aligned}

This is the required qua­dratic equa­tion. Its roots, found by fac­tori­sa­tion, are 25 and 30, and both make sense: check that 25×30=75025 \times 30 = 750 and 30×25=75030 \times 25 = 750.

A short his­tory

Qua­dratic equa­tions are very old. The Baby­lo­ni­ans are believed to have been the first to solve them: they could find two pos­i­tive num­bers from their sum and their prod­uct, which is the same as solv­ing x2−px+q=0x^2 - px + q = 0. Euclid gave a geo­met­ric way of find­ing lengths that, in today's lan­guage, are roots of qua­dratic equa­tions. Brah­magupta (598–665 CE) gave an explicit for­mula for equa­tions of the form ax2+bx=cax^2 + bx = c, and Srid­haracharya (about 1025 CE) derived the for­mula now known as the qua­dratic for­mula. The Arab math­e­mati­cian Al-Khwarizmi (about 800 CE) stud­ied qua­dratic equa­tions of dif­fer­ent types, and Abra­ham bar Hiyya Ha-Nasi, in his book Liber embado­rum pub­lished in Europe in 1145 CE, gave com­plete solu­tions of dif­fer­ent qua­dratic equa­tions.

Com­mon mis­takes

  • Decid­ing an equa­tion is qua­dratic because it con­tains x2x^2 some­where, with­out sim­pli­fy­ing. If the x2x^2 terms can­cel, as in Exam­ple 2, it is not qua­dratic.
  • Decid­ing an equa­tion is not qua­dratic because it con­tains x3x^3. If the x3x^3 terms can­cel, as in Exam­ple 4, it may well be qua­dratic.
  • For­get­ting the con­di­tion a≠0a \neq 0 in the def­i­n­i­tion.
  • Sign errors when expand­ing a prod­uct such as (x−5)(40−x)(x-5)(40-x): expand term by term and write every sign.
  • Read­ing off aa, bb and cc before putting the equa­tion in stan­dard form, for exam­ple tak­ing a=4a = 4 in 4x−3x2+2=04x - 3x^2 + 2 = 0.
  • For­get­ting to say what the let­ter stands for, with units, in a word prob­lem.

Try these

  1. Is (x+3)2=x2+6(x+3)^2 = x^2 + 6 a qua­dratic equa­tion? Answer: No. It sim­pli­fies to 6x+3=06x + 3 = 0, which is lin­ear.
  2. Is x(x−4)+2=(2x−1)(x+1)x(x-4) + 2 = (2x-1)(x+1) a qua­dratic equa­tion? Answer: Yes. It sim­pli­fies to x2+5x−3=0x^2 + 5x - 3 = 0.
  3. Write 5−2x2+3x=05 - 2x^2 + 3x = 0 in stan­dard form and state aa, bb and cc. Answer: −2x2+3x+5=0-2x^2 + 3x + 5 = 0, so a=−2a = -2, b=3b = 3, c=5c = 5 (equiv­a­lently 2x2−3x−5=02x^2 - 3x - 5 = 0).
  4. The prod­uct of two con­sec­u­tive pos­i­tive even inte­gers is 168. Form the qua­dratic equa­tion, tak­ing the smaller inte­ger as xx. Answer: x2+2x−168=0x^2 + 2x - 168 = 0.
  5. The sum of a non-zero num­ber and its rec­i­p­ro­cal is 103\displaystyle \frac{10}{3}. Form the qua­dratic equa­tion. Answer: 3x2−10x+3=03x^2 - 10x + 3 = 0.
  6. Is (x−1)3=x3+x(x-1)^3 = x^3 + x a qua­dratic equa­tion? Answer: Yes. It sim­pli­fies to −3x2+2x−1=0-3x^2 + 2x - 1 = 0, that is, 3x2−2x+1=03x^2 - 2x + 1 = 0.

Key terms

Qua­dratic equa­tion
An equa­tion that can be writ­ten as ax2+bx+c=0ax^2 + bx + c = 0 with aa, bb, cc real and a≠0a \neq 0.
Stan­dard form
The arrange­ment ax2+bx+c=0ax^2 + bx + c = 0, with terms in descend­ing pow­ers of xx and zero on the right.
Coef­fi­cient of x2x^2
The num­ber aa mul­ti­ply­ing x2x^2; it must not be zero.
Coef­fi­cient of xx
The num­ber bb mul­ti­ply­ing xx; it may be zero.
Con­stant term
The num­ber cc that does not mul­ti­ply any power of xx.
Degree
The high­est power of the vari­able that remains after the equa­tion is sim­pli­fied.
Root
A value of xx that makes the equa­tion true. A qua­dratic equa­tion has at most two roots.
Lin­ear equa­tion
An equa­tion of degree 1, such as x+12=0x + 12 = 0.

Com­mon ques­tions

Can bb or cc be zero in a qua­dratic equa­tion?

Yes. Only aa must be non-zero. For exam­ple, x2+7=0x^2 + 7 = 0 has b=0b = 0, and 2x2−5x=02x^2 - 5x = 0 has c=0c = 0; both are qua­dratic.

Is an equa­tion with x3x^3 in it ever qua­dratic?

It can be, if the x3x^3 terms can­cel when you sim­plify. The equa­tion (x+2)3=x3−4(x+2)^3 = x^3 - 4 reduces to x2+2x+2=0x^2 + 2x + 2 = 0, which is qua­dratic.

Why must the lead­ing coef­fi­cient be non-zero?

If a=0a = 0, the x2x^2 term van­ishes and the equa­tion becomes bx+c=0bx + c = 0, which is lin­ear. The con­di­tion keeps the squared term present.

Does mul­ti­ply­ing by −1-1 change the equa­tion?

No. Mul­ti­ply­ing every term by the same non-zero num­ber gives an equiv­a­lent equa­tion with exactly the same roots, so −x2+45x−324=0-x^2 + 45x - 324 = 0 and x2−45x+324=0x^2 - 45x + 324 = 0 have the same solu­tions.

Do I have to solve the equa­tion when a ques­tion says "rep­re­sent the sit­u­a­tion math­e­mat­i­cally"?

No. Such a ques­tion only asks for the equa­tion in stan­dard form. Solv­ing it is the next step, cov­ered in the les­son on fac­tori­sa­tion.

Ref­er­ences

  1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
  2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
  3. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.