Q:

98 points! Easy! Show work!

Accepted Solution

A:
1) Remember that the area of a square can be calculated using the equation [tex]A= s^{2} [/tex], where s=the length of one side of the square. You know that the area of the square is: A=[tex]25 y^{2} -20y+4[/tex]. Put that into the equation, and solve for s, the length of the side of the square by factoring:
[tex]A= s^{2}\\ 25 y^{2} -20y+4 = s^{2}\\ (5y-2)(5y-2)=s^{2}\\ (5y-2)^{2}=s^{2}\\ s= 5y-2[/tex]

The length of a side of the square is 5y-2.

2) To factor [tex]7 a^{2}-63 b^{2} [/tex],
First ask yourself, is there a greatest common factor of the coefficients? Yes, both 7 and 63 are divisible by 7 and 7 is the largest number that divides them both.

Factor out the 7: 
[tex]7 a^{2}-63 b^{2}\\ 7(a^{2}-9 b^{2})[/tex]

Now ask yourself, can [tex]7(a^{2}-9 b^{2}[/tex] be simplified any further? Yes. Remember your factoring rules for the difference of squares: [tex]x^{2} - y^{2} = (x+y)(x-y)[/tex]. For [tex]a^{2}-9 b^{2}[/tex], x = a and y = 3b. That means [tex]7(a^{2}-9 b^{2}) = 7(a+3b)(a-3b)[/tex]

Your final factored expression is: 7(a+3b)(a-3b)

3) When we are taking the perfect squares out of the radical, we are finding the simplest radical form for each radical. 
You're given: [tex] \sqrt{7}- \sqrt{24} + \sqrt{175} + \sqrt{150} [/tex]

To find simplest radical form, you must first prime factorize each number under the radical. That means taking the number breaking it down by prime factors so that its written as many prime factors multiplied together. Doing this will help you see which factors are squared, letting you take them out of the radical easier:
[tex]\sqrt{7} - \sqrt{24} + \sqrt{175} + \sqrt{150}\\ = \sqrt{7} - \sqrt{2*3*4} + \sqrt{5*5*7} + \sqrt{2*3*5*5}[/tex]

Next take out the number pairs under the radical, and put one of that number from the pair outside the radical (see picture for example).
You can see that 7 and 24 have no paired prime factors, so they don't have perfect squares that divide into them. That means they are in their simplest radical form, so leave them as they are in the expression.
175 breaks down into 5*5*7. There is a pair of fives, so you can take out the fives and put a 5 on the outside of the radical: [tex]\sqrt{175} = \sqrt{5*5*7} = 5 \sqrt{7}[/tex]
150 breaks down into 2*3*5*5. It also has a pair of fives, so you can take out the fives and put a 5 on the outside of the radical: [tex]\sqrt{150} = \sqrt{2*3*5*5} = 5 \sqrt{2*3} = 5 \sqrt{6}[/tex]

Next, put it all together into one expression:
[tex]\sqrt{7} - \sqrt{24} + \sqrt{175} + \sqrt{150}\\ = \sqrt{7} - \sqrt{24} + 5
\sqrt{7} + 5 \sqrt{6} [/tex]

Finally, simplify the expression. You can add or subtract radicals that have the same thing under the radical:
[tex]\sqrt{7} - \sqrt{24} + 5 \sqrt{7} + 5 \sqrt{6} \\ =6 \sqrt{7} - \sqrt{24} + 5 \sqrt{6}[/tex]

Your final answer is 6√7 - √24 + 5√6.

4) The perimeter of a square is calculated by adding up all the sides of the square. Since there are four sides on a square and each square is the same length, the equation [tex]P=4s[/tex] can be used to find the perimeter, where s=length of each side.

You're given the perimeter, p = [tex]28 a^{2} b^{4} [/tex], so plug it into the equation for the perimeter of a square to find s, the length of one side:
[tex]P=4s\\ 28 a^{2} b^{4} = 4s\\ s= 7a^{2} b^{4}[/tex]

Now you know the length of each side of the square, s = [tex]7a^{2} b^{4}[/tex]. Remember that the equation for the area of a square is: [tex]A=s^{2}[/tex], where s=length of the side of a square.

Since you know s = [tex]7a^{2} b^{4}[/tex], plug that into the equation for the area of a square and solve for A:
[tex]A=s^{2}\\ A={(7a^{2} b^{4})}^{2}\\ A = 7^{2} {(a^{2})}^{2} {(b^{4})}^{2} [/tex]

Remember your rules for exponents. When exponents are being raised to an exponent, you multiply the exponents: 
[tex]A = 7^{2} {(a^{2})}^{2} {(b^{4})}^{2}\\ A = 49 a^{4} b^{8} [/tex]

Your final answer for the area of the square is [tex]A = 49 a^{4} b^{8}[/tex]