SOLUTION: one of the questions to my homewok that i am having trouble with answering was r^2-3r/5+2/25=0 and the homework asks to find the value of the discriminant and describe the nature o

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: one of the questions to my homewok that i am having trouble with answering was r^2-3r/5+2/25=0 and the homework asks to find the value of the discriminant and describe the nature o      Log On


   

Question 686643: one of the questions to my homewok that i am having trouble with answering was r^2-3r/5+2/25=0 and the homework asks to find the value of the discriminant and describe the nature of the roots of each quadratic equation Then solve the equation and ive tried various ways but i have no confidence so o=if you can help that would be great
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
When you have the equation in the form
+y+=+a%2Ax%5E2+%2B+b%2Ax+%2B+c+, the the discriminant is
+b%5E2+-+4%2Aa%2Ac+
The rules are:
(1) If +b%5E2+-+4%2Aa%2Ac+%3C+0+, then there are 2 imaginary roots
(2) If +b%5E2+-+4%2Aa%2Ac+%3E+0+, then there are 2 real roots
(3) If +b%5E2+-+4%2Aa%2Ac+=+0+, then there is a single real root ( called a double root )
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Your equation is:
+f%28r%29+=+r%5E2+-+%283%2F5%29%2Ar+%2B+2%2F25+
+r%5E2+-+%283%2F5%29%2Ar+%2B+2%2F25+=+0+
( note that this is saying +f%28r%29+=+0+, which means that the
values of r that you find will be the x-crossings which are the roots )
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The fact that you have r and not x doesn't matter
The form is still +a%2Ax%5E2+%2B+b%2Ax+%2B+c+=+0+
+a+=+1+
+b+=+3%2F5+
+c+=+2%2F25+
Now just plug these into the discriminant
+b%5E2+-+4%2Aa%2Ac+=+%283%2F5%29%5E2+-+4%2A1%2A%282%2F25%29+
+b%5E2+-+4%2Aa%2Ac+=+9%2F25+-+8%2F25+
+b%5E2+-+4%2Aa%2Ac+=+1%2F25+
This is a positive result, so there are 2 roots, and they are real
Now to solve equation:
+r%5E2+-+%283%2F5%29%2Ar+%2B+2%2F25+=+0+
+%28+r+-+2%2F5+%29%2A%28+r+-+1%2F5+%29+=+0+ ( I could see the result by inspection )
+r+=+2%2F5+ and
+r+=+1%2F5+ are the solutions
But here's solution by completing the square:
+r%5E2+-+%283%2F5%29%2Ar+=+-2%2F25+
+r%5E2+-+%283%2F5%29%2Ar+%2B+%283%2F10%29%5E2+=+-2%2F25+%2B+%283%2F10%29%5E2+
+r%5E2+-+%283%2F5%29%2Ar+%2B+9%2F100+=+-2%2F25+%2B+9%2F100+
+r%5E2+-+%283%2F5%29%2Ar+%2B+9%2F100+=+-8%2F100+%2B+9%2F100+
+r%5E2+-+%283%2F5%29%2Ar+%2B+9%2F100+=+1%2F100+
+%28+r+-+3%2F10+%29%5E2+=+%281%2F10%29%5E2+
take the square root of both sides
+r+-+3%2F10+=+1%2F10+
+r+=+4%2F10+
+r+=+2%2F5+
But, also I can say that
+r+-+3%2F10+=+-1%2F10+
Taking the negative square root of (1/10)^2
+r+=+3%2F10+-+1%2F10+
+r+=+2%2F10+
+r+=+1%2F5+
I get the same answers
I hope this helps