I have a number of problems I'm having problem with but they start with the example given in the book
2x^2 - 10x
=2(x^2 - 5x)
=2{[x-(5/2)]^2 - [(5/2)^2]}
=2[x-(5/2)]^2 - (25/2)
Why isn't (5/2)^2 (25/4)?
I'm struggling with the exercises as well.
3x^2 - 24x
I get 3(x-4)^2 - 16 or if I try to follow the example, 3(x-4)^2 - 32, but the answer at the back of the book is 3(x-4)^2 - 48
I meant, "Why isn't (25/2)^2 (25/4)?"
You seem to have forgotten a basic rule of multiplying fractions.
a/b * c/d = ac/(bd)
The square of any fraction a/b is a^2/b^2.
For example, 2/3 of 2/3 = (2/3)^2 = 4/9
and (25/2)^2 = 625/4
Looking at your second problem,
3x^2 - 24x
= 3(x^2 -8x) Now add and subtract 48
= 3(x^2 -8x +16) -48
= 3(x-4)^2 -48
To understand why the expression (5/2)^2 is not equal to 25/4 in the given example, we need to clarify the order of operations involved.
In the example, we have the expression (5/2)^2. To evaluate this, we should follow the order of operations, commonly known as PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction).
According to PEMDAS, exponentiation is performed before multiplication and division. Therefore, we need to square the value 5/2 first.
(5/2)^2 = (5/2) * (5/2) = 25/4
As you correctly pointed out, (5/2)^2 is indeed equal to 25/4, and not 25/2.
Moving on to your second example, let's solve it step by step:
3x^2 - 24x
The first step is to factor out the greatest common factor, which is 3:
3(x^2 - 8x)
Next, we need to complete the square by adding and subtracting the square of half the coefficient of the x term. In this case, the coefficient is -8, so half of it is -4:
3(x^2 - 8x + (-4)^2 - (-4)^2)
Simplifying this expression gives us:
3(x - 4)^2 - 3(-4)^2
Note that when we subtract a square, we need to include a negative sign. Now, let's compute -3(-4)^2:
-3(-4)^2 = -3(16) = -48
The final expression becomes:
3(x - 4)^2 - 48
Therefore, the correct answer to the exercise is 3(x - 4)^2 - 48, matching the solution in the book.
Remember to double-check the calculations and carefully follow the order of operations when dealing with expressions and equations.