# Quadratic Equations - Problems (1)

This is a tutorial on using the quadratic equations to solve problems. Detailed solutions and explanations are included.

 Example - 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. x2 + y2 = 102 Solve the equation x + y + 10 = 24 for y. y = 14 - x Substitute y in the equation x2 + y2 = 102 by the expression obtained above. x2 + (14 - x)2 = 102 Expand the square, group like terms and write the above equation with the right side equal to zero. 2x2 -28x + 96 = 0 Multiply all terms in the above equation by 1/2. x2 -14x + 48 = 0 Find the discriminant of the above quadratic equation. Discriminant D = b2 - 4*a*c = 196 - 192 = 4 Use the quadratic formulas to solve the quadratic equation; two solutions x1 = [ -b + sqrt(D) ] / 2*a = [ 14 + 2 ] / 2 = 8 x2 = [ -b - sqrt(D) ] / 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 = sqrt (x2 + y2) = sqrt (82 cm2 + 62 cm2) = sqrt(64 cm2 + 36 cm2) = 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 m2. Find the length x and width y, x > y, of the rectangle. Example - 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. x2 + (x + 1)2 = 61 Expand (x + 1)2, group like terms and write the above equation with the right side equal to zero. 2x2 + 2x - 60 = 0 Multiply all terms in the above equation by 1/2. x2 + x - 30 = 0 Find the discriminant of the above quadratic equation. Discriminant D = b2 - 4*a*c = 1 + 120 = 121 Use the quadratic formulas to solve the quadratic equation; two solutions x1 = [ -b + sqrt(D) ] / 2*a = [ -1 + 11 ] / 2 = 5 x2 = [ -b - sqrt(D) ] / 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: 52 + 62 = 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. More references and links on how to Solve Equations, Systems of Equations and Inequalities.