A lin­ear equa­tion such as 3x+5=113x + 5 = 11 has the unknown only to the first power, and it has exactly one solu­tion. When the unknown appears squared, as in x2−5x+6=0x^2 - 5x + 6 = 0, the equa­tion is called qua­dratic, and it can have two solu­tions. Qua­dratic equa­tions appear when­ever an area, a prod­uct of two unknowns, or a speed and a time are involved. This les­son sets out what a qua­dratic equa­tion is, how to write it in stan­dard form, and what is meant by its roots.

Stan­dard form

An equa­tion of the form

ax2+bx+c=0ax^2 + bx + c = 0

where aa, bb, cc are real num­bers and a≠0a \neq 0, is called a qua­dratic equa­tion in xx. This is its stan­dard form. The num­bers aa, bb and cc are called the coef­fi­cients: aa is the coef­fi­cient of x2x^2, bb the coef­fi­cient of xx, and cc the con­stant term.

The con­di­tion a≠0a \neq 0 mat­ters: if a=0a = 0, the x2x^2 term dis­ap­pears and the equa­tion is lin­ear, not qua­dratic. But bb or cc may be zero.

Equa­tion aa bb cc
x2−5x+6=0x^2 - 5x + 6 = 0 11 −5-5 66
2x2+7x−4=02x^2 + 7x - 4 = 0 22 77 −4-4
x2−9=0x^2 - 9 = 0 11 00 −9-9
3x2+12x=03x^2 + 12x = 0 33 1212 00

Bring­ing an equa­tion to stan­dard form

Many equa­tions are qua­dratic in dis­guise. Expand any brack­ets, clear frac­tions, and col­lect every term on one side.

Exam­ple 1

Write x(x−2)=5x(x - 2) = 5 in stan­dard form.

Expand: x2−2x=5x^2 - 2x = 5. Move 5 to the left: x2−2x−5=0x^2 - 2x - 5 = 0. So a=1a = 1, b=−2b = -2, c=−5c = -5.

Exam­ple 2

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

Expand the left side: x2+2x+1=x2+3x^2 + 2x + 1 = x^2 + 3. The x2x^2 terms can­cel, leav­ing 2x−2=02x - 2 = 0, which is lin­ear. So it is not a qua­dratic equa­tion, even though it con­tains x2x^2. Always sim­plify before decid­ing.

Exam­ple 3

Write x2+3x=2\displaystyle \frac{x}{2} + \frac{3}{x} = 2 in stan­dard form, where x≠0x \neq 0.

Mul­ti­ply every term by 2x2x, the com­mon denom­i­na­tor: x2+6=4xx^2 + 6 = 4x. Col­lect terms: x2−4x+6=0x^2 - 4x + 6 = 0.

Exam­ple 4

Is (x−2)(x+3)=(x+1)(2x−1)(x - 2)(x + 3) = (x + 1)(2x - 1) qua­dratic?

Left: x2+x−6x^2 + x - 6. Right: 2x2+x−12x^2 + x - 1. So x2+x−6=2x2+x−1x^2 + x - 6 = 2x^2 + x - 1, which gives 0=x2+50 = x^2 + 5, that is x2+5=0x^2 + 5 = 0. This is qua­dratic, with a=1a = 1, b=0b = 0, c=5c = 5.

Roots of a qua­dratic equa­tion

A num­ber is called a root (or a solu­tion) of the equa­tion ax2+bx+c=0ax^2 + bx + c = 0 if putting it in place of xx makes the left side equal to 0. To solve the equa­tion means to find all its roots.

A qua­dratic equa­tion has at most two real roots. It may have two dif­fer­ent roots, one repeated root, or no real root at all. We will see how to tell which in a later les­son.

Exam­ple 5

Check whether x=2x = 2 and x=3x = 3 are roots of x2−5x+6=0x^2 - 5x + 6 = 0. Is x=1x = 1 a root?

For x=2x = 2: 4−10+6=04 - 10 + 6 = 0. Yes, 2 is a root.

For x=3x = 3: 9−15+6=09 - 15 + 6 = 0. Yes, 3 is a root.

For x=1x = 1: 1−5+6=21 - 5 + 6 = 2, not 0. So 1 is not a root.

Exam­ple 6

If x=2x = 2 is a root of x2+kx−10=0x^2 + kx - 10 = 0, find kk.

Since 2 is a root, putting x=2x = 2 must give 0:

4+2k−10=0  ⟹  2k=6  ⟹  k=34 + 2k - 10 = 0 \implies 2k = 6 \implies k = 3

Exam­ple 7

If x=−12\displaystyle x = -\frac{1}{2} is a root of 2x2+px−3=02x^2 + px - 3 = 0, find pp.

2×14−p2−3=0  ⟹  12−p2=3  ⟹  1−p=6  ⟹  p=−5\displaystyle 2 \times \frac{1}{4} - \frac{p}{2} - 3 = 0 \implies \frac{1}{2} - \frac{p}{2} = 3 \implies 1 - p = 6 \implies p = -5

From words to a qua­dratic equa­tion

Exam­ple 8

The prod­uct of two con­sec­u­tive nat­ural num­bers is 132. Write a qua­dratic equa­tion for the smaller num­ber.

Let the smaller num­ber be xx. The next num­ber is x+1x + 1. Their prod­uct is x(x+1)=132x(x + 1) = 132, so

x2+x−132=0x^2 + x - 132 = 0

You can check that x=11x = 11 is a root: 121+11−132=0121 + 11 - 132 = 0. So the num­bers are 11 and 12.

Exam­ple 9

The length of a rec­tan­gu­lar plot in Vijayawada is 3 m more than its width, and its area is 108 square metres. Write a qua­dratic equa­tion for the width.

Let the width be xx m. The length is x+3x + 3 m, and the area is x(x+3)=108x(x + 3) = 108, so x2+3x−108=0x^2 + 3x - 108 = 0. (We will solve it in the next les­son: the width is 9 m.)

Prac­tice

  1. Write the val­ues of aa, bb, cc for 5x2−3x+8=05x^2 - 3x + 8 = 0 and for 7−x2=07 - x^2 = 0.
  2. Write (x−3)(x+4)=10(x - 3)(x + 4) = 10 in stan­dard form.
  3. Write 2x2=3(x+5)2x^2 = 3(x + 5) in stan­dard form.
  4. Is x2+4x=(x+2)2x^2 + 4x = (x + 2)^2 a qua­dratic equa­tion? Explain.
  5. Is x(2x+1)=x2+6x(2x + 1) = x^2 + 6 a qua­dratic equa­tion? If so, write it in stan­dard form.
  6. Write x+1x=3\displaystyle x + \frac{1}{x} = 3 in stan­dard form, where x≠0x \neq 0.
  7. Check whether x=−1x = -1 and x=4x = 4 are roots of x2−3x−4=0x^2 - 3x - 4 = 0.
  8. If x=3x = 3 is a root of x2−7x+k=0x^2 - 7x + k = 0, find kk.
  9. If x=23\displaystyle x = \frac{2}{3} is a root of 3x2+mx+2=03x^2 + mx + 2 = 0, find mm.
  10. The sum of a num­ber and its square is 90. Write a qua­dratic equa­tion for the num­ber.

Answers

Show answers
  1. a=5a = 5, b=−3b = -3, c=8c = 8; and a=−1a = -1, b=0b = 0, c=7c = 7 (or, mul­ti­ply­ing by −1-1, x2−7=0x^2 - 7 = 0 with a=1a = 1, c=−7c = -7).
  2. x2+x−12=10x^2 + x - 12 = 10, so x2+x−22=0x^2 + x - 22 = 0.
  3. 2x2−3x−15=02x^2 - 3x - 15 = 0
  4. No: the right side is x2+4x+4x^2 + 4x + 4, so all the terms in xx can­cel and the equa­tion says that 0 equals 4, which is never true. It is not a qua­dratic equa­tion and has no solu­tion.
  5. Yes: 2x2+x=x2+62x^2 + x = x^2 + 6 gives x2+x−6=0x^2 + x - 6 = 0.
  6. x2−3x+1=0x^2 - 3x + 1 = 0
  7. x=−1x = -1: 1+3−4=01 + 3 - 4 = 0, a root. x=4x = 4: 16−12−4=016 - 12 - 4 = 0, a root.
  8. 9−21+k=09 - 21 + k = 0, so k=12k = 12.
  9. 3×49+2m3+2=0\displaystyle 3 \times \frac{4}{9} + \frac{2m}{3} + 2 = 0 gives 43+2m3+2=0\displaystyle \frac{4}{3} + \frac{2m}{3} + 2 = 0, so 4+2m+6=04 + 2m + 6 = 0 and m=−5m = -5.
  10. x2+x−90=0x^2 + x - 90 = 0 (the roots are 9 and −10-10).