28
Solving word problems with quadratic equations
Textbook: pp. 163–166
GoalTranslate a word problem into a quadratic equation, solve it and check the roots against the conditions of the problem.
New words
choosing the unknown · noma’lumni belgilashforming an equation · tenglama tuzishroot that fits the condition · shartga mos ildizchecking the answer · javobni tekshirish
Explanation
Solving a word problem with a quadratic equation has four stages. First, the unknown quantity is denoted by x and other quantities are expressed through x. Second, an equation is formed from the conditions (area, perimeter, speed, time, sum and product of numbers and so on). Third, the quadratic equation is solved. Fourth, the roots are checked against the problem: lengths, times and speeds must be positive, and numbers of people or objects must be natural numbers, so sometimes one root does not fit and is discarded. The answer is written with units.
Worked examples
A rectangle is 3 cm longer than it is wide and its area is 40 cm². Let the width be x cm: x(x + 3) = 40, x² + 3x - 40 = 0, (x + 8)(x - 5) = 0. x = -8 does not fit, x = 5. Width 5 cm, length 8 cm.
A cyclist rode 30 km at v km/h. With a speed 5 km/h higher he would take 1 hour less: 30/v - 30/(v + 5) = 1, v² + 5v - 150 = 0, (v + 15)(v - 10) = 0, v = 10 km/h. Check: 30 : 10 = 3 hours, 30 : 15 = 2 hours.
Class activity
“Problem lab”: groups invent their own word problems (area, journey, numbers); another group solves them with an equation and checks the roots.
Practice
1
The product of two consecutive natural numbers is 56. Find the numbers.
7 and 8
2
A rectangle has perimeter 26 cm and area 40 cm². Find its sides.
5 cm and 8 cm
3
The legs of a right triangle differ by 7 and the hypotenuse is 13. Find the legs.
5 and 12
4
Why is a negative root often discarded in a word problem?
Because lengths, times and speeds cannot be negative; the root may be correct algebraically but does not fit the conditions of the problem.