Quadratic Equations - Word Questions
E.g.1
The sum of two numbers is 27 and their product is 50. Find the numbers.
Let one number be x. Then the other number is 50/x.
x + 50/x = 27
X x => x
2 + 50 = 27x
- 27x => x
2 - 27x + 50 = 0
(x -25)(x -2) = 0
(x -25) = 0 or (x -2) = 0
x = 25 or x = 2.
E.g.2
The length of a rectangle is 5 cm more than its width and the area is 50cm
2. Find the length, width and the perimeter.
Let the width be x. Then the length = x + 5.
x(x + 5) = 50
x
2 + 5x = 50
-50 => x
2 + 5x - 50 = 0
(x + 10)(x -5) =0
(x + 10) = 0 or (x -5) =0
x = -10 or x = 5 - x = -10 is impossible to be a width
Width = 5cm; so, the length = 10cm.
Perimeter = 30cm.
E.g.3
The three sides of a right-angled triangle are x, x+1 and 5. Find x and the area, if the longest side is 5.
The hypotenuse = 5
x
2 + (x+1)
2 = 5
2 (Pythagoras' Theorem)
x
2 + x
2 + 2x + 1 = 25
-25 => x
2 + x
2 + 2x - 24 = 0
(x + 6)(x - 4) = 0
(x + 6) = 0 or (x - 4) = 0
x = -6 or x = 4
x = 4;
Area = 0.5 x 3 x 4 = 6cm
2
E.g.4
The product of two numbers is 24 and the mean is 5. Find the numbers.
Let one number = x; then the other = 24/x
(x + 24/x)/2 = 5
X 2 => x + 24/x = 10
X x => x
2 + 24 = 10x
- 10x => x
2 + -10x + 24 = 0
(x - 6)(x -4) = 0
(x - 6) = 0 or (x -4) = 0
x = 6 or x =4
The numbers are 6 or 4.
E.g.5
The sum of numbers is 9. The squares of the numbers is 41. Find the numbers.
These are
quadratic simultaneous equations.
let the numbers be x and y.
x + y = 9
x
2 + y
2 = 41
From the first equation, y = (9-x)
Now substitute this in the second equation.
x
2 + (9-x)
2 = 41
x
2 + 81 - 18x + x
2 = 41
2x
2 - 16x + 81 = 41
2x
2 - 16x + 40 = 0
x
2 - 8x + 20 = 0
(x - 5)(x -4) =0
(x - 5) = 0 or (x -4) =0
x = 5 or x = 4
Substitute in the first equation, y = 5 or 4
The numbers are 5 and 4.