The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two...

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Transcript of The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two...

Page 1: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.
Page 2: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.
Page 3: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.
Page 4: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.

The Principle of Square Roots

Let’s consider x2 = 25. We know that the number 25 has two real-number square roots, 5 and -5, the solutions of the equation. Thus we see that square roots can provide quick solutions for equations of the type x2 = k, where k is a constant.

Page 5: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.
Page 6: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.

Example

Solution

Solve (x + 3)2 = 7

2( 3) 7x

3 7 or 3 7x x

The solutions are 3 7.

3 7 or 3 7.x x

Page 7: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.

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Completing the Square

Not all quadratic equations can be solved as in the previous examples. By using a method called completing the square, we can use the principle of square roots to solve any quadratic equation.

Solve x2 + 10x + 4 = 0

Page 8: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.

Example

Solution

Solve x2 + 10x + 4 = 0

2( 5) 21x

2 10 4 0x x

2 10 4x x

x2 + 10x + 25 = –4 + 25

5 21 or 5 21x x

5 21. The solutions are

Using the principle of square roots

Factoring

Adding 25 to both sides.

Page 9: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.
Page 10: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.
Page 11: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.

2.591,0 0.257,0

Page 12: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.

Example

Solution

Jackson invested $5800 at an interest rate of r, compounded annually. In 2 years, it grew to $6765. What was the interest rate?

1. Introduction. We are already familiar with the compound-interest formula.

(1 )tA P r

6765 = 5800(1 + r)2

The translation consists of substituting into the formula:

Page 13: The Principle of Square Roots Let’s consider x 2 = 25. We know that the number 25 has two real-number square roots, 5 and 5, the solutions of the.

2. Body. Solve for r:

6765/5800 = (1 + r)2

6765/5800 1 r

1 6765/5800 r

.080 or 2.080r r

Since the interest rate cannot be negative, the solution is .080 or 8.0%.

3. Conclusion. The interest rate was 8.0%.


8 Math Square and Square Roots

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