Solving Word Problems Involving Quadratic Equations
Solving Word Problems Involving Quadratic Equation
One of the most important topics in Algebra is solving quadratic equations. It has found wide applications in many areas of Mathematics. This hub presents several problems involving quadratic equations with their solution. Hope you will enjoy this.
Sample Problem One :
The sum of two numbers is 18 and the sum of their square is 170. Find the numbers.
Solution:
Let X = the first number
18 – X = the second number
X^2 = square of the first number
(18-X)^2 = square of the second number
Working Equation:
X^2 + (18-X)^2 = 170
X^2 + 324 -36X + X^2 = 170
2X^2 -36X +324 -170 = 0
(2X^2 -36X + 154 = 0)1/2
X^2 – 18X + 77 = 0
Factoring X^2 -18X + 77 = 0
(X – 7) (X – 11) = 0
X – 7 = 0 therefore X = 7
X – 11 = 0 therefore X = 7
The two number are 7 and 11.
Problem Number Two :
One leg of a right triangle is 7 cm shorter than the other leg. Its area is 30cm^2. Find its perimeter.
Solution :
Let X = length of one leg
X + 7 = length of the other leg
Area of a triangle = (bh)/2
Working Equation :
X (X+7)/2 = 30
(X(X+7)/2 = 30) 2
X^2 + 7X = 60
X^2 + 7X - 60 = 0
Factoring X^2 + 7X -60
(X -5 ) ( X + 12 ) = 0
X – 5 = 0 therefore X = 5
Leg1 = 5
Leg2 = 5 + 7 = 12
Finding the hypotenuse:
Hypotenuse = SQRT(5^2 + 12^2) = SQRT(25 + 144) = SQRT (169) =13
Perimeter = 5+12+13 = 30 vcentimeters.
Problem Number Three :
One side of a triangle is one dm. less than its diagonal. If the perimeter of the rectangle is 62, find the area of the rectangle.
Solution :
Length of the rectangle = d – 1
To represent width use phytagorean theorem and consider diagonal as hypotenuse:
d^2 = w^2 + (d – 1)^2
Rearranging the equation : w^2 = d^2 - (d-1)^2
w^2 = d^2 –d^2 +2d – 1
w^2 = 2d -1
w = SQRT(2d -1)
Working Equation : Based from Perimeter = 2l + 2W
2(d-1) + 2 SQRT(2d- 1 ) = 62
2d – 2 + 2 SQRT(2d-1) = 62
(2SQRT(2d -1 ) = 64 -2d) ½
SQRT(2d – 1) = 32 – d
(SQRT(2d -1 ))^2 = (32 –d)^2
2d -1 = 1024 -64d + d^2
d^2 - 66d + 1025 = 0
Factoring gives (d – 41) (d -25) = 0
d – 25 = 0
d = 25
Length = d – 1 = 25 – 1 = 24
Width = SQRT(2d-1) =SQRT(50- 1) = SQRT49 = 7
Area of the rectangle = (24)(7) = 168 dm^2