Forming Quadratic Equations

Area and product problems become quadratics.

Name the unknown

A rectangle is 2 cm longer than it is wide, and its area is 24 cm². How wide is it?

Start by giving the unknown a letter. Let the width be x cm. The length is 2 cm more than the width, so it is (x + 2) cm. Now both sides of the rectangle are written in terms of one letter.

x + 2x

The shorter side, the width, is x. The longer side, the length, is x + 2.

Write the fact as an equation

The area of a rectangle is its length times its width, so x(x + 2) = 24. Expand the bracket: x × x = x² and x × 2 = 2x, so x² + 2x = 24.

The equation has an x² term, so it is a quadratic equation. Multiplying two lengths that both contain x is what makes the x² appear: an area, or any product of two unknown amounts, often leads to a quadratic.

x²2xx2xarea 24

Split the length into x and 2. The two pieces have areas x × x = x² and x × 2 = 2x, and together they make the whole area, 24.

Solve it, then read the answer in the story

To solve a quadratic equation by factoring, one side must be 0. Subtract 24 from both sides: x² + 2x − 24 = 0.

Look for two numbers that multiply to −24 and add to 2. They are 6 and −4, so (x + 6)(x − 4) = 0. A product is 0 only when one of its factors is 0, so x = −6 or x = 4.

Both numbers solve the equation, but x is a width, and a width cannot be negative. So x = −6 is rejected. The rectangle is 4 cm wide and 4 + 2 = 6 cm long. Check: 4 × 6 = 24.

Sometimes both roots are answers

Two consecutive integers multiply to 132. Let the smaller one be n; the next integer is n + 1. Their product is n(n + 1) = 132, which is n² + n − 132 = 0.

The numbers 12 and −11 multiply to −132 and add to 1, so (n + 12)(n − 11) = 0, and n = −12 or n = 11. Integers may be negative, so this time both roots give an answer: 11 and 12, or −12 and −11. Check: 11 × 12 = 132 and (−12) × (−11) = 132.

So reject a root only when the story rules it out, and say why.

Forming an equation from its roots

You can also build a quadratic equation backwards, from the roots it must have. For roots 2 and −5, the brackets are (x − 2) and (x + 5), since each is 0 at one root. So (x − 2)(x + 5) = 0, which expands to x² + 3x − 10 = 0.

The same equation comes from the sum and product of the roots: x² − (sum)x + (product) = 0. Here the sum is 2 + (−5) = −3 and the product is 2 × (−5) = −10, so x² − (−3)x + (−10) = x² + 3x − 10.

The usual mistakes

Writing x = 24 or x + 2 = 24 straight from x(x + 2) = 24. Only a product of 0 forces one of its factors to be 0, so rearrange to x² + 2x − 24 = 0 before you factor.

Expanding x(x + 2) as x² + 2. The x in front multiplies every term inside the bracket, so the second term is 2x.

Giving both roots as the answer without reading the story. A length, a time or a number of people cannot be negative.

Worked example: A Rectangular Patio with a Known Area and a Length 3 m More Than Its Width

Question A rectangular patio is 3 m longer than it is wide, and its area is 40 m2. (a) Find the width and the length of the patio. (b) An edging strip runs all the way round the patio. How long is the strip?

  1. 1.Let the width be x m. The length is 3 m more, so it is (x + 3) m. The area is length times width, so x(x + 3) = 40.

    area 40 m2xx + 3x(x + 3)=40
    area 40 m2xx + 3x(x + 3)=40
    The width is x m and the length is (x + 3) m, so the area gives x(x + 3) = 40.

  2. 2.Expand the bracket: x2 + 3x = 40. Subtract 40 from both sides, so that one side is zero: x2 + 3x − 40 = 0.

    area 40 m2xx + 3x(x + 3)=40x2+ 3x=40expand the bracketx2+ 3x − 40=0subtract 40 from both sides
    area 40 m2xx + 3x(x + 3)=40expand the bracketx2+ 3x=40subtract 40 from both sidesx2+ 3x − 40=0
    Expand the bracket and subtract 40 from both sides: x2 + 3x − 40 = 0.

  3. 3.Factorize. Two numbers with a product of −40 and a sum of 3 are 8 and −5, so (x + 8)(x − 5) = 0.

    area 40 m2xx + 3x(x + 3)=40x2+ 3x=40expand the bracketx2+ 3x − 40=0subtract 40 from both sides(x + 8)(x − 5)=0factorize
    area 40 m2xx + 3x(x + 3)=40expand the bracketx2+ 3x=40subtract 40 from both sidesx2+ 3x − 40=0factorize(x + 8)(x − 5)=0
    The numbers 8 and −5 have a product of −40 and a sum of 3: (x + 8)(x − 5) = 0.

  4. 4.A product is zero only when one of its factors is zero, so x = −8 or x = 5. A width cannot be negative, so x = −8 is rejected. (a) The width is 5 m and the length is 5 + 3 = 8 m. Check: 5 × 8 = 40.

    area 40 m2x = 5 mx + 3 = 8 mx(x + 3)=40x2+ 3x=40expand the bracketx2+ 3x − 40=0subtract 40 from both sides(x + 8)(x − 5)=0factorizex = −8 or x = 5x = −8 is rejected: a width cannot be negativex = 5, and 5 × 8 = 40
    area 40 m2x = 5 mx + 3 = 8 mx(x + 3)=40expand the bracketx2+ 3x=40subtract 40 from both sidesx2+ 3x − 40=0factorize(x + 8)(x − 5)=0x = −8 or x = 5x = −8 is rejected: a width cannot be negativex = 5, and 5 × 8 = 40
    (a) x = −8 or x = 5. A width cannot be negative, so the width is 5 m and the length is 8 m.

  5. 5.(b) The strip is the perimeter of the patio: 2 × (5 + 8) = 26 m.

    area 40 m2x = 5 mx + 3 = 8 mx(x + 3)=40x2+ 3x=40expand the bracketx2+ 3x − 40=0subtract 40 from both sides(x + 8)(x − 5)=0factorizex = −8 or x = 5x = −8 is rejected: a width cannot be negativex = 5, and 5 × 8 = 40perimeter = 2 × (5 + 8) = 26 m
    area 40 m2x = 5 mx + 3 = 8 mx(x + 3)=40expand the bracketx2+ 3x=40subtract 40 from both sidesx2+ 3x − 40=0factorize(x + 8)(x − 5)=0x = −8 or x = 5x = −8 is rejected: a width cannot be negativex = 5, and 5 × 8 = 40perimeter = 2 × (5 + 8) = 26 m
    (b) The strip is 2 × (5 + 8) = 26 m long.

Answer: (a) The width is 5 m and the length is 8 m; (b) 26 m

Common mistakes

  • Solving x(x + 3) = 40 by writing x = 40 or x + 3 = 40. Only a product of zero forces one of its factors to be zero, so the equation must be rearranged to x2 + 3x − 40 = 0 before it is factorized.
  • Giving both x = −8 and x = 5 as widths. Both numbers satisfy the equation, but x is a length in meters, and a length cannot be negative.

More quadratic equations problems, worked step by step →

Pythagoras' theorem gives a quadratic too

The next application is a right-angled triangle. Its longest side, opposite the right angle, is called the hypotenuse. Pythagoras' theorem says that the squares of the two shorter sides add up to the square of the hypotenuse. For a triangle with sides 3, 4 and 5: 3² + 4² = 9 + 16 = 25 = 5².

When a side is written with x, squaring it makes an x² term, so the equation from the theorem is a quadratic equation.

Worked example: A Right-Angled Shade Sail with Edges x and x + 7 and a Known Longest Edge

Question A shade sail is a right-angled triangle. The two edges that meet at the right angle are x m and (x + 7) m long, and the longest edge is 13 m long. (a) Find x and the lengths of the two shorter edges. (b) Find the area of the sail.

  1. 1.By Pythagoras' theorem, the squares of the two shorter edges add up to the square of the hypotenuse: x2 + (x + 7)2 = 132.

    xx + 713 mx2+ (x + 7)2=132
    xx + 713 mx2+ (x + 7)2=132
    By Pythagoras' theorem, x2 + (x + 7)2 = 132.

  2. 2.Expand: (x + 7)2 = x2 + 14x + 49 and 132 = 169, so 2x2 + 14x + 49 = 169.

    xx + 713 mx2+ (x + 7)2=1322x2+ 14x + 49=169expand both squares
    xx + 713 mx2+ (x + 7)2=132expand both squares2x2+ 14x + 49=169
    Expand both squares: 2x2 + 14x + 49 = 169.

  3. 3.Subtract 169 from both sides: 2x2 + 14x − 120 = 0. Divide both sides by 2: x2 + 7x − 60 = 0.

    xx + 713 mx2+ (x + 7)2=1322x2+ 14x + 49=169expand both squares2x2+ 14x − 120=0subtract 169 from both sidesx2+ 7x − 60=0divide both sides by 2
    xx + 713 mx2+ (x + 7)2=132expand both squares2x2+ 14x + 49=169subtract 169 from both sides2x2+ 14x − 120=0divide both sides by 2x2+ 7x − 60=0
    Subtract 169 from both sides, then divide both sides by 2: x2 + 7x − 60 = 0.

  4. 4.Factorize. Two numbers with a product of −60 and a sum of 7 are 12 and −5, so (x + 12)(x − 5) = 0, which gives x = −12 or x = 5. A length cannot be negative, so x = −12 is rejected. (a) x = 5, and the edges are 5 m and 5 + 7 = 12 m long. Check: 52 + 122 = 25 + 144 = 169 = 132.

    x = 5 mx + 7 = 12 m13 mx2+ (x + 7)2=1322x2+ 14x + 49=169expand both squares2x2+ 14x − 120=0subtract 169 from both sidesx2+ 7x − 60=0divide both sides by 2(x + 12)(x − 5)=0factorizex = −12 or x = 5x = −12 is rejected: a length cannot be negativex = 5, and 25 + 144 = 169
    x = 5 mx + 7 = 12 m13 mx2+ (x + 7)2=132expand both squares2x2+ 14x + 49=169subtract 169 from both sides2x2+ 14x − 120=0divide both sides by 2x2+ 7x − 60=0factorize(x + 12)(x − 5)=0x = −12 or x = 5x = −12 is rejected: a length cannot be negativex = 5, and 25 + 144 = 169
    (a) (x + 12)(x − 5) = 0. A length cannot be negative, so x = 5, and the edges are 5 m and 12 m long.

  5. 5.(b) The two shorter edges meet at the right angle, so one is the base and the other is the height: the area is 12 × 5 × 12 = 30 m2.

    x = 5 mx + 7 = 12 m13 mx2+ (x + 7)2=1322x2+ 14x + 49=169expand both squares2x2+ 14x − 120=0subtract 169 from both sidesx2+ 7x − 60=0divide both sides by 2(x + 12)(x − 5)=0factorizex = −12 or x = 5x = −12 is rejected: a length cannot be negativex = 5, and 25 + 144 = 169area = 1/2 × 5 × 12 = 30 m2
    x = 5 mx + 7 = 12 m13 mx2+ (x + 7)2=132expand both squares2x2+ 14x + 49=169subtract 169 from both sides2x2+ 14x − 120=0divide both sides by 2x2+ 7x − 60=0factorize(x + 12)(x − 5)=0x = −12 or x = 5x = −12 is rejected: a length cannot be negativex = 5, and 25 + 144 = 169area = 1/2 × 5 × 12 = 30 m2
    (b) The area of the sail is 12 × 5 × 12 = 30 m2.

Answer: (a) x = 5, so the edges are 5 m and 12 m long; (b) 30 m2

Common mistakes

  • Expanding (x + 7)2 as x2 + 49. The bracket is multiplied by itself, (x + 7)(x + 7), and that gives the middle term 14x as well.
  • Writing x + (x + 7) = 13. Pythagoras' theorem relates the squares of the sides, not the sides themselves, and in any triangle the two shorter sides add up to more than the longest side.

More quadratic equations problems, worked step by step →

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