SOLUTION: D..ex I have no idea on how to do these types of problems Problem #7 Solve by completing the square x^2=5x+2 Problem #8 Find the x-intercepts y= x^2+5x+2 Problem #9

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: D..ex I have no idea on how to do these types of problems Problem #7 Solve by completing the square x^2=5x+2 Problem #8 Find the x-intercepts y= x^2+5x+2 Problem #9      Log On


   

Question 78077: D..ex
I have no idea on how to do these types of problems
Problem #7
Solve by completing the square
x^2=5x+2
Problem #8
Find the x-intercepts
y= x^2+5x+2
Problem #9
Is this a trinomial a perfect square? Explain why or why not.
x^2+18x+81

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
#7
x%5E2=5x%2B2
x%5E2-5x-2=0 Get all terms to one side
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=1+x%5E2-5+x-2 Start with the given equation



y%2B2=1+x%5E2-5+x Add 2 to both sides



y%2B2=1%28x%5E2-5x%29 Factor out the leading coefficient 1



Take half of the x coefficient -5 to get -5%2F2 (ie %281%2F2%29%28-5%29=-5%2F2).


Now square -5%2F2 to get 25%2F4 (ie %28-5%2F2%29%5E2=%28-5%2F2%29%28-5%2F2%29=25%2F4)





y%2B2=1%28x%5E2-5x%2B25%2F4-25%2F4%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 25%2F4 does not change the equation




y%2B2=1%28%28x-5%2F2%29%5E2-25%2F4%29 Now factor x%5E2-5x%2B25%2F4 to get %28x-5%2F2%29%5E2



y%2B2=1%28x-5%2F2%29%5E2-1%2825%2F4%29 Distribute



y%2B2=1%28x-5%2F2%29%5E2-25%2F4 Multiply



y=1%28x-5%2F2%29%5E2-25%2F4-2 Now add %2B2 to both sides to isolate y



y=1%28x-5%2F2%29%5E2-33%2F4 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=1, h=5%2F2, and k=-33%2F4. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=1x%5E2-5x-2 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1x%5E2-5x-2%29 Graph of y=1x%5E2-5x-2. Notice how the vertex is (5%2F2,-33%2F4).



Notice if we graph the final equation y=1%28x-5%2F2%29%5E2-33%2F4 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C1%28x-5%2F2%29%5E2-33%2F4%29 Graph of y=1%28x-5%2F2%29%5E2-33%2F4. Notice how the vertex is also (5%2F2,-33%2F4).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.





So we get:
%28x-2.5%29%5E2-8.25=0
%28x-2.5%29%5E2=8.25 Add 8.25 to both sides
x-2.5=0%2B-sqrt%288.25%29 Take the square root of both sides
x=0%2B-sqrt%288.25%29%2B2.5 Add 2.5 to both sides
x=0%2B-sqrt%288.25%29%2B2.5
So we get:
x=sqrt%288.25%29%2B2.5 or x=-sqrt%288.25%29%2B2.5
Which is approximately
x=5.3723 or x=-0.3723

#8
To find the x intercepts we can use the quadratic formula:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B5x%2B2+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%285%29%5E2-4%2A1%2A2=17.

Discriminant d=17 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+17+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285%29%2Bsqrt%28+17+%29%29%2F2%5C1+=+-0.43844718719117
x%5B2%5D+=+%28-%285%29-sqrt%28+17+%29%29%2F2%5C1+=+-4.56155281280883

Quadratic expression 1x%5E2%2B5x%2B2 can be factored:
1x%5E2%2B5x%2B2+=+1%28x--0.43844718719117%29%2A%28x--4.56155281280883%29
Again, the answer is: -0.43844718719117, -4.56155281280883. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5%2Ax%2B2+%29


So our x-intercepts are
x=-4.56155281280883 and x=-0.43844718719117

#9
The trinomial is a perfect square since it can be written as
%28x%2B9%29%5E2
Which is a representation of a squared number. For instance if we let x=1 we get
%281%2B9%29%5E2=%2810%29%5E2=100
And we can see that 100 is a perfect square. Notice we can let x=1 for this equation:
%281%29%5E2%2B18%281%29%2B81=1%2B18%2B81=19%2B81=100 which gives us the same result.
So since the expression x%5E2%2B18x%2B81 generates perfect squares for any whole x's (make a table and you'll see it), it is a perfect square