1

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Solution: This equation has fractions in it. How do we clear the fractions? Multiply all of the terms by the LCD, n(n + 4). Then we solve using an appropriate method. Also, we must note which values of n will cause division by 0, and exclude those from our solution.

n 0 and n 4

26(n+4) 12n n +4n

n n+4 n n+46 12

n n

n n+4 14

2 6n+24 12n n +4n 224 6n n +4n

2 0 n +4n+6n 24 2 0 n 10n 24

0 = (n + 12)(n – 2)

Answer: {– 12, 2}

So far, we have worked with 3 different methods for solving quadratic equations. We have solved by: factoring, completing the square, and using the quadratic formula. In this section we will look at two types of equations and their methods of solutions.

6Solv

12 1

n n

4e.Example 1.

Answer: 15,3

Your Turn Problem #1

9 16Solve: 1

x x 5

2

7.5 More Quadratic Equations

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Solution: Although this is not a quadratic (2nd degree) equation, it is in quadratic form. You may recall this equation is still factorable if we can find two numbers whose product is the last number and whose difference is the middle number. Since we have x4, we will have x2 in both first positions of the parentheses. Also, this can still be factored by grouping if you wish.

Example 2. Solve: x4 + 5x2 – 84 = 0

x 2i 3 or x = 7 2i 3, 7

4 2x 5x 84 0

2 2x 12 x 7 0

2 2x 12 0 or x 7 0

2 2x 12 or x 7

x 12 or x 7

or by grouping,

4 2x 5x 84 0

1 84 84 1 84 2 42 3 28 4 21 6 14 7 12

4 2 2x 7x 12x 84 0

2 2 2x x 7 12 x 7 0

2 2x 7 x 12 0

Answer: i 10, 2 2

Your Turn Problem #2

4 2Solve: x 2x 80 0

3

7.5 More Quadratic Equations

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Applications

“Applications” means word problems. Here’s a reminder of steps to take when solving word problems.

1. Choose a meaningful variable to represent the unknown quantity. Represent any other unknowns in terms of that variable.

2. Sketch any figure, chart, or diagram that might be helpful in analyzing the problem if possible.

3. Look for a guideline that you can use to set up an equation. Is there a formula we can use? Are we translating? Etc.

4. Set up the equation.

5. Solve the equation.

Next Slide

6. Answer the question.

4

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Let x = the height of the tower.Step 1. Name the unknown.

16 yards

xx + 4

Step 2. A diagram is useful in this problem; draw it. Since the distance form the top of the tower to the point 16 yards from the base is 4 more than the height, we will name the hypotenuse x +4.

Step 3. The guideline. Notice that we have a right triangle. Therefore we can use the Pythagorean Theorem. (i.e., a2+b2=c2.)

Step 4. The equation.Step 5. Solve.

Answer: The tower is 30 yards high.

At a point 16 yards from the base of a tower, the distance to the top of the tower is 4 yards more than the height of the tower. Find the height of the tower.Solution:

Example 3.

22 216 x x 4 2 2256 x x 8x 16

256 8x 16 x 30

A 15-foot ladder is leaned against a wall. If the distance from the base of the ladder to the wall is 3 feet, how far up the wall does the ladder reach? Approximate your answer to the nearest tenth of a foot.

Your Turn Problem #3

Answer: The ladder reaches up approximately 14.7 feet.The End.B.R.6-3-07