Word Problems and Practical Applications

The final page of the exam focuses on Question 6, which presents two word problems applying quadratic equations to real-world scenarios.

a) The first problem involves finding a number that, when decreased by 5 and multiplied by itself plus 2, equals 408. This is solved by setting up the equation (x - 5)(x + 2) = 408 and solving it to find x = 22.

b) The second problem deals with calculating the dimensions of a cuboid given its volume and height, and a relationship between its length and width. The problem is solved step-by-step:

• Volume: 765 cm³
• Height: 17 cm
• Length = Width + 4 cm

Example: The solution involves setting up the equation (y + 4) · y · 17 = 765, where y represents the width of the cuboid. Solving this quadratic equation yields y = 5 cm (width) and x = 9 cm (length).

This question demonstrates the practical application of quadratische Gleichungen Aufgaben (quadratic equation problems) in real-world contexts, reinforcing the importance of these mathematical concepts.

Question Details and Solutions

This page provides detailed solutions for the first four questions of the exam.

Question 1 focuses on determining the Definitionsmenge Wurzeln (definition set for roots). The solutions show that for √(7x-1), the definition set is x ≥ 1/7, while for √(-4x-6), it's x ≤ -1.5.

Question 2 involves simplifying root expressions and rationalizing denominators. Notable solutions include:

• √225x² simplified to 15√x²
• √7p² √49p² √7p simplified to 14p√p

Example: For rationalizing the denominator of 3/(√6 - √2), the solution multiplies both numerator and denominator by (√6 + √2), resulting in 3(√6 + √2)/(6 - 2) = 3(√6 + √2)/4.

Question 3 requires students to identify a quadratic equation from a graph, with the solution given as x² + 0.5x - 2 = 0.

Question 4 involves solving a quadratic equation using the completing the square method. The equation 2x² - 4.8x + 0.88 = 0 is solved step-by-step, resulting in solutions x = 0.2 and x = 2.2.

Advanced Equation Solving

This page continues with Question 5, which involves solving various types of quadratic and related equations.

a) 6x² = 6144 is solved to yield x = ±32. b) x² - 6x - 7 = 0 is solved using the quadratic formula, resulting in x = 7 or x = -1. c) 3x² + 17 = -(x + 1)(x - 1) is transformed and solved, showing that it has no real solutions. d) (2x-9)(5x+3) = 0 is solved using the zero product property, giving x = 4.5 or x = -0.6.

Highlight: The solution notes that using the zero product property for part d) would have been simpler than expanding the equation.

e) -3x² - 4x = 0 is factored and solved, resulting in x = 0 or x = -4/3.

These problems demonstrate various techniques for solving quadratische Gleichungen (quadratic equations), including factoring, using the quadratic formula, and applying algebraic properties.