To determine the number of real zeros of f(x)=x^2-8x+16, we will solve the equation f(x)=0 by .
f(x)=0 ⇔ x^2-8x+16=0
Then we can compare this method to finding the number of real zeros by using the .
Factoring
To solve the equation by factoring, we will start by identifying the values of a, b, and c.
x^2-8x+16=0 ⇕ 1x^2+( - 8)x+ 16=0We have a with a= 1, b= - 8, and c= 16. To factor the left-hand side, we need to find a factor pair of 1 * 16=16 whose sum is - 8.
Since 16 is a positive number, we will only consider with the same sign — both positive or both negative — so that their product is positive.
| Factor Pair
|
Product of Factors
|
Sum of Factors
|
| 1 and 16
|
^(1* 16) 16
|
1+16 17
|
| - 1 and - 16
|
^(- 1* (- 16)) 16
|
- 1+(- 16) - 17
|
| 2 and 8
|
^(2* 8) 16
|
2+8 10
|
| - 2 and - 8
|
^(- 2* (- 8)) 16
|
- 2+(- 8) - 10
|
| 4 and 4
|
^(4* 4) 16
|
4+4 8
|
| - 4 and - 4
|
^(- 4* (- 4)) 16
|
- 4+(- 4) - 8
|
The whose product is 16 and whose sum is - 8 are - 4 and - 4. With this information, we can rewrite the linear factor on the left-hand side of the equation, and .
x^2-8x+16=0
x^2-4x-4x+16=0
x(x-4)-4x+16=0
x(x-4)-4(x-4)=0
(x-4)(x-4)=0
Now we can solve for x.
We found that the only unique solution to the given equation is x=4. Therefore, the number of real zeros of f(x)=x^2-8x+16 is one.
Discriminant
The discriminant is the expression under the radical sign in the , b^2-4ac. Let's analyze the Quadratic Formula.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a
We know that the discriminant can be used to determine the number of real solutions of a quadratic equation by looking at the sign of the result.
| Discriminant
|
Number of Real Solutions
|
| Negative
|
None
|
| Zero
|
One
|
| Positive
|
Two
|
In our case, a= 1, b= - 8, and c= 16. Let's find the discriminant.
b^2-4 a c
â–¼
Substitute values and evaluate
( - 8)^2-4( 1)( 16)
64-4(1)(16)
64-64
0
Since the discriminant is 0, the number of real zeros of f(x)=x^2-8x+16 is one.
Comparison
Let's look at the similarities and differences between the methods.
- Similarities.
- Both methods are algebraic.
- We transform algebraic expressions, substitute, and perform various types of transformations.
- Differences.
- The first method is much longer and requires more mathematical calculations.
- The second method only requires evaluating the discriminant and checking its sign.
