Quadratic Equations  Problems (1)
Using quadratic equations to solve problems; detailed solutions and explanations are included.
 Problems with Solutions
Problem 1:
A right triangle has a perimeter of 24 cm and a hypotenuse of 10 cm. Find the sides x and y, x > y, that make the right angle of the triangle.Solution to Problem 1:

We start by drawing a triangle with the given information

The perimeter of the triangle is 24, hence
x + y + 10 = 24

It is a right triangle, use Pythagoras theorem to obtain.
x^{2} + y^{2} = 10^{2}

Solve the equation x + y + 10 = 24 for y.
y = 14  x

Substitute y in the equation x^{2} + y^{2} = 10^{2} by the expression obtained above.
x^{2} + (14  x)^{2} = 10^{2}

Expand the square, group like terms and write the above equation with the right side equal to zero.
2x^{2} 28x + 96 = 0

Multiply all terms in the above equation by 1/2.
x^{2} 14x + 48 = 0

Find the discriminant of the above quadratic equation.
Discriminant Δ = b^{2}  4*a*c = 196  192 = 4

Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = [ b + √Δ ] / (2 a) = [ 14 + 2 ] / 2 = 8
x2 = [ b  √Δ ] / (2 a) = [ 14  2 ] / 2 = 6

use the equation y = 14  x to find the corresponding value of y.
y1 = 14  8 = 6
y2 = 14  6 = 8

Taking into account the condition x > y, the sides that make the right angle of the triangle are: x = 8 cm and y = 6 cm.
Check answer:
Hypotenuse h = √ (x^{2} + y^{2})
= √ (8^{2} cm^{2} + 6^{2} cm^{2})
= √(64 cm^{2} + 36 cm^{2})
= 10 cm, it agrees with the given value.
Perimeter = y + x + hypotenuse
= 8 cm + 6 cm + 10 cm
= 24 cm, it agrees with the given value.
Matched Problem 1:
A rectangle has a perimeter of 60 m and an area of 200 m^{2}. Find the length x and width y, x > y, of the rectangle.
Solutions to Matched Problems
Problem 2:
The sum of the squares of two consecutive real numbers is 61. Find the numbers.Solution to Problem 2:

Let x and x+1 be the two consecutive numbers. The sum of the square of x and x + 1 is equal to 61.
x^{2} + (x + 1)^{2} = 61

Expand (x + 1)^{2}, group like terms and write the above equation with the right side equal to zero.
2x^{2} + 2x  60 = 0

Multiply all terms in the above equation by 1/2.
x^{2} + x  30 = 0

Find the discriminant of the above quadratic equation.
Discriminant Δ = b^{2}  4*a*c = 1 + 120 = 121

Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = [  b + √Δ ] / 2*a = [ 1 + 11 ] / 2 = 5
x2 = [  b  √Δ ] / 2*a = [ 1  11 ] / 2 = 6

First solution to the problem
first number: x1 = 5
second number: x1 + 1 = 6

Second solution to the problem
first number: x2 = 6
second number: x2 + 1 = 5
Check answer:
first solution sum of squares: 5^{2} + 6^{2}
= 25 + 36 = 61
second solution sum of squares: (6)^{2} + (5)^{2}
= 36 + 25 = 61
The two solutions to the problem agree with the given information in the problem.
Matched Problem 2:
The sum of the squares of two consecutive even real numbers is 52. Find the numbers.
Solutions to Matched Problems
Solutions to Matched ProblemsSolution to Matched Problem 1:

The perimeter of the rectangle is 60 m, hence
2 x + 2 y = 60

The area of the rectangle is 200 m^{2}, hence
x y = 200

Solve the equation 2 x + 2 y = 60 for y.
y = 30  x

Substitute y in the equation x y = 200 by the expression for y obtained above.
x(30  x) = 200

Multiply, group like terms and write the above equation with the right hand side equal to zero.
 x^{2} +30 x  200 = 0

Find the discriminant of the above quadratic equation.
Discriminant D = b^{2}  4 a c = 900  800 = 100

Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = ( b + √D ) / (2 a) = (30 + 10 ) / ( 2) = 10 m
x2 = ( b  √D ) / (2 a) = (30  10 ) / ( 2) = 20 m

use y = 30  x found above to find the corresponding value of y.
y1 = 30  10 = 20 m
y2 = 30  20 = 10 m
 Taking into account the condition x > y, the length x = 20 m and the width y = 10 m.
As an exercise, check the perimeter and the area.
Solution to Matched Problem 2:

Let x and x + 2 (the difference between two consecutive even numbers is 2) be the two consecutive even numbers. The sum of the square of x and x + 2 is equal to 52, hence
x^{2} + (x + 2)^{2} = 52

Expand (x + 2)^{2}, group like terms and write the above equation with the right side equal to zero.
2x^{2} + 4x  48 = 0

Multiply all terms in the above equation by 1/2 to obtain the following equivalent equation.
x^{2} + 2 x  24 = 0

Find the discriminant of the above quadratic equation.
Discriminant D = b^{2}  4 a c = 4 + 90 = 100

Use the quadratic formulas to solve the quadratic equation; two solutions
x1 = (  b + √D ) / (2 a) = (  2 + 10 ) / 2 = 4
x2 = (  b  √D ) / (2 a) = (  2  10 ) / 2 =  6

First solution to the problem
first number: x1 = 4
second number: x1 + 2 = 6

Second solution to the problem
first number: x2 =  6
second number: x2 + 2 =  4
As an exercise check that the square of the two numbers, for each solution, is 52.
More References and linksQuadratic Functions Problems with Solutions
Parabola Questions and Problems with Detailed Solutions
Solve Quadratic Equations Using Discriminants
Solve Equations, Systems of Equations and Inequalities.
