Applications: Quadratic Equations

The area of a rectangle is its length multiplied by its width. With the width as the unknown, the area gives an equation in which the unknown is multiplied by itself, which is a quadratic equation. It is solved by bringing every term to one side and factorizing.

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.
2. Expand the bracket: x² + 3x = 40. Subtract 40 from both sides, so that one side is zero: x² + 3x − 40 = 0.
3. Factorize. Two numbers with a product of −40 and a sum of 3 are 8 and −5, so (x + 8)(x − 5) = 0.
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.
5. (b) The strip is the perimeter of the patio: 2 × (5 + 8) = 26 m.

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

techniqueForming Quadratic Equations · Solving by Factoring

examsO-Level · SAT · GCSE Higher

Pythagoras' theorem says that in a right-angled triangle the square of the longest side, the hypotenuse, is equal to the sum of the squares of the other two sides. Here both shorter sides are written in terms of x, so the theorem gives a quadratic equation in x.

1. By Pythagoras' theorem, the squares of the two shorter edges add up to the square of the hypotenuse: x² + (x + 7)² = 13².
2. Expand: (x + 7)² = x² + 14x + 49 and 13² = 169, so 2x² + 14x + 49 = 169.
3. Subtract 169 from both sides: 2x² + 14x − 120 = 0. Divide both sides by 2: x² + 7x − 60 = 0.
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.
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.

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

techniqueForming Quadratic Equations · Solving by Factoring

examsO-Level · SAT · GCSE Higher

The ball is on the ground whenever its height is zero, so the landing time is a root of 20m − 5m² = 0. The greatest height is read from the completed-square form of the formula: a square is never negative, so the height is largest when the squared bracket is zero.

1. The ball is on the ground when h = 0, so 20m − 5m² = 0. Both terms have the factor 5m: 5m(4 − m) = 0.
2. One of the factors is zero, so m = 0 or m = 4. The root m = 0 is the moment the ball is thrown. (a) The ball lands after 4 seconds. Check: 20 × 4 − 5 × 4² = 80 − 80 = 0.
3. Take out the factor −5: h = −5(m² − 4m). Half of 4 is 2, and (m − 2)² = m² − 4m + 4, so m² − 4m = (m − 2)² − 4.
4. Substitute this into the bracket: h = −5[(m − 2)² − 4] = 20 − 5(m − 2)².
5. The square (m − 2)² is never negative, so h is never more than 20, and h = 20 when m = 2. (b) The greatest height is 20 meters, reached 2 seconds after the ball is thrown. Check: at m = 1 and at m = 3 the height is 15 meters, which is lower.

answer(a) After 4 seconds; (b) 20 meters, reached 2 seconds after the ball is thrown

techniqueCompleting the Square · Solving by Factoring · Sketching from Vertex and Factored Form

examsO-Level · SAT · GCSE Higher

The frame adds its width to both ends of each side, so each outer side is the side of the picture plus twice the width. The area of the outer rectangle gives a quadratic equation. Its roots are not whole numbers, so it is solved by the quadratic formula: the roots of ax² + bx + c = 0 are x = −b ± √b² − 4ac2a.

1. Let the width of the frame be x cm. The frame adds x cm at both ends of each side, so the outer rectangle is (30 + 2x) cm by (20 + 2x) cm, and (20 + 2x)(30 + 2x) = 900.
2. Expand: 600 + 40x + 60x + 4x² = 900, which is 4x² + 100x + 600 = 900. Subtract 900 from both sides: 4x² + 100x − 300 = 0. Divide both sides by 4: x² + 25x − 75 = 0.
3. No two whole numbers have a product of −75 and a sum of 25, so use the quadratic formula x = −b ± √b² − 4ac2a with a = 1, b = 25 and c = −75. First, b² − 4ac = 625 + 300 = 925.
4. So x = −25 ± √9252, and √925 ≈ 30.414. This gives x ≈ 2.707 or x ≈ −27.707. A width cannot be negative, so the second root is rejected. (a) The frame is 2.71 cm wide, correct to 2 decimal places.
5. (b) Keep the unrounded width. The outer sides are 30 + 2 × 2.707 = 35.414 cm and 20 + 2 × 2.707 = 25.414 cm, so the line is 2 × (35.414 + 25.414) = 121.656 cm long, which is 122 cm to the nearest centimeter.

answer(a) 2.71 cm; (b) 122 cm

techniqueThe Quadratic Formula · Forming Quadratic Equations

examsO-Level · SAT · GCSE Higher

A quadratic equation with roots α and β is (x − α)(x − β) = 0, which expands to x² − (α + β)x + αβ = 0. So when the sum and the product of two numbers are known, the equation that has them as roots can be written down at once: x² − (sum)x + (product) = 0.

1. Let the two ages be the roots of a quadratic equation. The sum of the roots is 19 and their product is 84, so the equation is x² − 19x + 84 = 0.
2. Factorize. The product 84 is positive and the sum is −19, so both numbers are negative: they are −7 and −12, and (x − 7)(x − 12) = 0.
3. So x = 7 or x = 12, and both roots are used, one for each sister. (a) The equation is x² − 19x + 84 = 0, and the sisters are 7 and 12 years old. Check: 7 + 12 = 19 and 7 × 12 = 84.
4. Let it be k years ago. Then the sisters were (12 − k) and (7 − k) years old, so 12 − k = 2(7 − k). Expand the bracket: 12 − k = 14 − 2k. Add 2k to both sides: 12 + k = 14. Subtract 12 from both sides: k = 2.
5. (b) It was 2 years ago, when the sisters were 10 and 5 years old. Check: 10 = 2 × 5.

answer(a) x² − 19x + 84 = 0; the sisters are 7 and 12 years old; (b) 2 years ago

techniqueThe Sum and Product of the Roots · Solving by Factoring

examsSAT

The stall makes a profit when P is more than zero, which is a quadratic inequality. Its roots are the prices at which the profit is exactly zero, and the graph of P shows on which side of the roots the profit is positive. The graph is symmetrical, so its highest point is halfway between the roots.

1. The stall makes a profit when P > 0, so −5x² + 60x − 100 > 0. Divide both sides by −5. Dividing by a negative number reverses the inequality sign: x² − 12x + 20 < 0.
2. Factorize: two numbers with a product of 20 and a sum of −12 are −2 and −10, so (x − 2)(x − 10) < 0. The profit is exactly zero at the roots x = 2 and x = 10.
3. The coefficient of x² in P is negative, so the graph of P opens downward, and it is above the x-axis between the roots. Test a price in each part: at x = 6, P = −180 + 360 − 100 = 80, and at x = 1 and at x = 11, P = −45.
4. (a) The stall makes a profit when 2 < x < 10, that is, when a cup costs more than $2 and less than $10.
5. (b) The highest point of the graph is halfway between the roots, at x = 2 + 102 = 6, where P = 80. The greatest profit is $80 a day, at a price of $6 a cup.

answer(a) 2 < x < 10: a price of more than $2 and less than $10; (b) a price of $6, which gives a profit of $80

techniqueSolving a Quadratic Inequality · Sketching from Vertex and Factored Form

examsO-Level · GCSE Higher · H2

The width of a pen with a given area is a root of a quadratic equation. For ax² + bx + c = 0 the number b² − 4ac is called the discriminant. When it is positive the equation has two real roots, when it is zero it has one, and when it is negative it has none. If there is no real root, there is no width that works, so the pen cannot be made.

1. The perimeter is 40 m, so the length and the width add up to 20 m. Let the width be x m. Then the length is (20 − x) m, and an area of 120 m² needs x(20 − x) = 120.
2. Expand the bracket: 20x − x² = 120. Add x² to both sides and subtract 20x from both sides: x² − 20x + 120 = 0.
3. Here a = 1, b = −20 and c = 120, so the discriminant is b² − 4ac = (−20)² − 4 × 1 × 120 = 400 − 480 = −80. It is negative, so the equation has no real roots. (a) No width gives 120 m², so the pen cannot have that area.
4. The area of the pen is A = 20x − x² = −(x² − 20x). Half of 20 is 10, and (x − 10)² = x² − 20x + 100, so x² − 20x = (x − 10)² − 100 and A = 100 − (x − 10)².
5. The square (x − 10)² is never negative, so A is never more than 100, and A = 100 when x = 10. (b) The greatest area is 100 m², when the pen is a square of side 10 m. Check: a pen of 9 m by 11 m has an area of 99 m².

answer(a) No: the discriminant is −80, so no width gives an area of 120 m²; (b) 100 m², when the pen is a square of side 10 m

techniqueThe Discriminant and the Nature of the Roots · Completing the Square · Forming Quadratic Equations

examsO-Level · SAT · GCSE Higher

At a point where the beam meets the arch, the same x and y satisfy both equations. Substituting the expression for y from the line into the curve leaves a quadratic equation in x alone. Each root is the x-coordinate of one meeting point, and the line then gives its y-coordinate.

1. At a meeting point both equations hold, so the two expressions for y are equal: x + 4 = 6x − x².
2. Add x² to both sides and subtract 6x from both sides: x² − 5x + 4 = 0. Its discriminant is b² − 4ac = 25 − 16 = 9, which is positive, so there are two real roots and the beam meets the arch at two points.
3. Factorize: two numbers with a product of 4 and a sum of −5 are −1 and −4, so (x − 1)(x − 4) = 0, and x = 1 or x = 4.
4. Substitute each root into y = x + 4: y = 5 when x = 1, and y = 8 when x = 4. (a) The beam meets the arch at (1, 5) and (4, 8). Check in the curve: 6 × 1 − 1² = 5 and 6 × 4 − 4² = 8.
5. (b) From (1, 5) to (4, 8) the beam goes 3 m across and 3 m up. These are the two shorter sides of a right-angled triangle, so by Pythagoras' theorem the length is √3² + 3² = √18 ≈ 4.24 m.

answer(a) (1, 5) and (4, 8); (b) √18 ≈ 4.24 m

techniqueA Linear Equation Paired with a Quadratic · When a Line Meets, Touches or Misses a Curve

examsO-Level · SAT · GCSE Higher

Time is distance divided by speed. With the usual speed as the unknown, each of the two times is a fraction with the unknown in its denominator, and their difference is 1 hour. Multiplying both sides by both denominators clears the fractions and leaves a quadratic equation.

1. Let the usual speed be v km/h. Time is distance divided by speed, so the usual time is 300v hours and the faster time is 300v + 10 hours. The faster journey is 1 hour shorter: 300v − 300v + 10 = 1.
2. Multiply both sides by v(v + 10) to clear the fractions: 300(v + 10) − 300v = v(v + 10).
3. Expand: 300v + 3000 − 300v = v² + 10v, so 3000 = v² + 10v. Subtract 3000 from both sides: v² + 10v − 3000 = 0.
4. Factorize: two numbers with a product of −3000 and a sum of 10 are 60 and −50, so (v + 60)(v − 50) = 0, and v = −60 or v = 50. A speed cannot be negative, so v = −60 is rejected. (a) The usual speed is 50 km/h.
5. (b) The faster speed is 50 + 10 = 60 km/h, so the journey would take 30060 = 5 hours. Check: the usual time is 30050 = 6 hours, and 6 − 5 = 1 hour.

answer(a) 50 km/h; (b) 5 hours

techniqueFractional Equations · Solving by Factoring

examsO-Level · SAT · GCSE Higher

Completing the square writes the expression as a squared bracket plus a number. A squared bracket is never negative, so the whole expression is never less than that number. If the number is positive, the expression is always positive, and the number is its least value.

1. Half of 12 is 6, and (x − 6)² = x² − 12x + 36, so x² − 12x = (x − 6)² − 36.
2. Substitute this into the formula: C = (x − 6)² − 36 + 50 = (x − 6)² + 14.
3. The square (x − 6)² is never negative, so C is at least 14 for every value of x. (a) The cost is always positive, so it can never be zero. The discriminant agrees: b² − 4ac = 144 − 200 = −56 is negative, so x² − 12x + 50 = 0 has no real roots.
4. (b) C is least when (x − 6)² = 0, which is when x = 6. The least cost is $14 for each mug, when 600 mugs are made in a week. Check: at x = 5 and at x = 7 the cost is 1 + 14 = $15.

answer(a) C = (x − 6)² + 14, which is at least 14, so it is never zero; (b) $14, when 600 mugs are made in a week

techniqueAlways Positive or Always Negative Quadratics · Completing the Square · The Discriminant and the Nature of the Roots

examsO-Level · SAT

Completing the square turns x² + bx into a squared bracket minus a number, and a square root then undoes the square. It works when the equation does not factorize with whole numbers, and it works just as well with letters in place of the numbers. Made with letters, the same moves give one formula for every sheet: this is how the quadratic formula itself is found from ax² + bx + c = 0.

1. Let the margin be x cm. It runs along both ends of each side, so the printed rectangle is (60 − 2x) cm by (40 − 2x) cm. The sheet covers 60 × 40 = 2400 cm², and half of it is 1200 cm², so (60 − 2x)(40 − 2x) = 1200.
2. Expand: 2400 − 120x − 80x + 4x² = 1200, which is 4x² − 200x + 2400 = 1200. Subtract 1200 from both sides and divide both sides by 4: x² − 50x + 300 = 0. Then move the constant across: x² − 50x = −300.
3. Complete the square. Half of 50 is 25, and (x − 25)² = x² − 50x + 625, so add 625 to both sides: (x − 25)² = 325. Take the square root of both sides, with both signs: x − 25 = ±√325, so x = 25 ± √325, and √325 ≈ 18.028.
4. This gives x ≈ 6.972 or x ≈ 43.028. Two margins of 43.028 cm cannot fit across a sheet only 40 cm wide, so that root is rejected. (a) The margin is 6.97 cm wide, correct to 2 decimal places. Check: 60 − 2x = 10 + 2√325 and 40 − 2x = 2√325 − 10, and their product is 4 × 325 − 100 = 1200.
5. (b) Make the same moves with letters. (L − 2x)(W − 2x) = A expands to 4x² − 2(L + W)x + LW − A = 0. Divide both sides by 4 and move the constant across: x² − L + W2x = A − LW4. Half of L + W2 is L + W4, so add its square, (L + W)²16, to both sides: (x − L + W4)² = (L + W)² − 4LW + 4A16 = (L − W)² + 4A16.
6. Take the square root of both sides: x = L + W ± √(L − W)² + 4A4. Because 4A is positive, the square root is more than W − L, so the root with + is more than L + W + W − L4 = W2, and two margins that wide are wider than the sheet. (b) The margin is x = L + W − √(L − W)² + 4A4, the root with the minus sign. Check: with L = 60, W = 40 and A = 1200 it gives 100 − √52004 ≈ 6.97 cm, as in (a).

answer(a) 6.97 cm; (b) x = L + W − √(L − W)² + 4A4, the root with the minus sign

techniqueDeriving the Quadratic Formula · Completing the Square · Forming Quadratic Equations

examsSAT · GCSE Higher

The braking distance sits under a square root in the formula for the speed. Getting the root alone on one side and squaring both sides gives a quadratic equation, but squaring also accepts a value that made the two sides equal in size and opposite in sign. So each root of the quadratic is checked in the equation before squaring, and a root that fails it is rejected.

1. Let the braking distance be d m. The car was traveling at 4√d m/s, so in the 1 second before the brakes went on it covered 4√d m. The two distances add up to 60 m: 4√d + d = 60.
2. Get the square root alone on one side: 4√d = 60 − d. Square both sides: 16d = 3600 − 120d + d².
3. Bring every term to one side: d² − 136d + 3600 = 0. Two numbers with a product of 3600 and a sum of −136 are −36 and −100, so (d − 36)(d − 100) = 0, and d = 36 or d = 100.
4. Squaring can bring in a root that the equation before squaring does not have, so check each one in 4√d = 60 − d. For d = 36: 4 × 6 = 24 and 60 − 36 = 24, which agree. For d = 100: 4 × 10 = 40, but 60 − 100 = −40. Squaring turned 40 = −40 into 1600 = 1600, so d = 100 is rejected; a braking distance of 100 m cannot fit inside a stop of 60 m in any case. (a) The braking distance was 36 m.
5. (b) The speed was v = 4√36 = 24 m/s. An hour has 3600 seconds and a kilometer has 1000 m, so this is 24 × 3.6 = 86.4 km/h. Yes: the car was over the limit of 80 km/h. Check: 24 m during the reaction and 36 m of braking make 60 m.

answer(a) 36 m; (b) yes: the car was traveling at 24 m/s, which is 86.4 km/h

techniqueSolving Equations with a Square Root · Solving by Factoring · Forming Quadratic Equations

examsO-Level · SAT

The time for the trip upstream is a fraction with the unknown speed in its denominator, and that denominator is negative for a boat slower than the current. Multiplying through by it would keep the inequality for some speeds and reverse it for others, so everything is brought to one side over one denominator and the sign of each factor is tested. The story then says which speeds are possible.

1. Let the boat's speed in still water be v km/h. Against the current it moves upstream at (v − 4) km/h, so the trip takes 12v − 4 hours. The boat makes headway only if v − 4 is positive, so v > 4. The trip takes less than 1 hour when 12v − 4 < 1.
2. The sign of v − 4 decides whether multiplying by it keeps or reverses the inequality, so do not multiply. Subtract 1 from both sides and write over one denominator: 12 − (v − 4)v − 4 < 0, which is 16 − vv − 4 < 0. The critical values are v = 16, where the top is zero, and v = 4, where the bottom is zero.
3. Test the sign of each factor. For v < 4: the top is positive and the bottom negative, so the quotient is negative. For 4 < v < 16: both are positive, so the quotient is positive. For v > 16: the top is negative and the bottom positive, so the quotient is negative. The inequality holds when v < 4 or v > 16.
4. Below 4 km/h the boat is carried downstream and never reaches the lake; the formula gives a negative time there, which is less than 1 but means nothing. Those speeds are rejected. (a) The trip upstream takes less than 1 hour when v > 16 km/h. Check: at v = 20 it takes 1216 = 0.75 hour.
5. (b) Downstream the boat moves at (v + 4) km/h, and 1 hour 15 minutes is 54 hours, so 12v − 4 + 12v + 4 < 54. Adding the fractions gives 24v(v − 4)(v + 4) < 54. Subtract 54 and write over one denominator: 96v − 5(v² − 16)4(v − 4)(v + 4) < 0, which is −5v² + 96v + 804(v − 4)(v + 4) < 0. Multiply both sides by −4, a negative number, and reverse the sign: (5v + 4)(v − 20)(v − 4)(v + 4) > 0.
6. For v > 4 the factors 5v + 4, v − 4 and v + 4 are all positive, so the quotient has the sign of v − 20, and it is positive when v > 20. (b) The round trip takes less than 1 hour 15 minutes when v > 20 km/h. Check: at v = 20 it takes 1216 + 1224 = 0.75 + 0.5 = 1.25 hours, exactly 1 hour 15 minutes, and at v = 24 it takes 0.6 + 1228 ≈ 1.03 hours.

answer(a) v > 16 km/h; (b) v > 20 km/h

techniqueSolving a Rational Inequality · Fractional Equations · Solving a Quadratic Inequality

examsSAT · H2