SOLUTION: Hello all! Thank God for the Matha Tutors!. Could someone please help me get with this one? Here it is: z^2 + 11/2z = -6, The answer in the book say its (-3/4,-4) Could some

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello all! Thank God for the Matha Tutors!. Could someone please help me get with this one? Here it is: z^2 + 11/2z = -6, The answer in the book say its (-3/4,-4) Could some      Log On


   

Question 17983: Hello all!
Thank God for the Matha Tutors!. Could someone please help me get with this one? Here it is:
z^2 + 11/2z = -6, The answer in the book say its (-3/4,-4)
Could someone explain this to me?
Thanks to all,
AC
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
z^2 + 11/2z = -6
z^2 + 11/2z + 6 = 0
Now this is a quadratic equation which can be solved as:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B5.5x%2B6+=+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=%285.5%29%5E2-4%2A1%2A6=6.25.

Discriminant d=6.25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5.5%2B-sqrt%28+6.25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285.5%29%2Bsqrt%28+6.25+%29%29%2F2%5C1+=+-1.5
x%5B2%5D+=+%28-%285.5%29-sqrt%28+6.25+%29%29%2F2%5C1+=+-4

Quadratic expression 1x%5E2%2B5.5x%2B6 can be factored:
1x%5E2%2B5.5x%2B6+=+%28x--1.5%29%2A%28x--4%29
Again, the answer is: -1.5, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5.5%2Ax%2B6+%29


Now the solution we get,as you see above,is different from what your book says.
Lets check according to the book.


Now the book says the roots are (-3/4 and -4).
then the expression would become
+%28x%2B%283%2F4%29%29%2A%28x%2B4%29=0+
+x%5E2%2B%283%2F4%29x%2B4x%2B%28%283%2F4%29%2A4%29=0+
+x%5E2+%2B+%284%2B%283%2F4%29%29x+%2B+3+=+0+
+x%5E2+%2B+4.75x+%2B+3+=+0+
+x%5E2+%2B+%2819%2F4%29x+%2B+3+=+0+
This is the equation you would need to obtain the roots given in your book.


Hope this helps,
Prabhat