other Quadratic equations
posted by theresa .
The problem is 7x^3x+1=x^3+3x^2+x. Find all real answers. I came upwith two answers 3x^2=1 and 2x=1/2. The first answer just does not look right. I must be forgeting something!
You can check them by putting in x=1/4 (the second answer), and then x=+ sqrt(1/3) These are three answers that need to be checked.
If the equation balances, the answers are correct.
The other way of checking it is to graph
y=7x^3x+1x^33x^2x. yes, combine the terms. Where y is zero, ie, crossing the x axis, that is a real solution.
Respond to this Question
Similar Questions

solving Equations
Solve x square pluse 5x subtracting by 6 equals zero? 
Please Check My Math Work
Hi.  I solved this problem,but I am not sure if it is correct.If you can please tell me if it is right or if it is wrong what is wrong with it.: 6y^254= (3y6)(2y+9) *Many of you may be are wondering why my title is,"[C]razy Baby. … 
math, correction plz
Can someone correct these for me.PLZ Problem#1 Directions solve equation ãx+4 = 3 My answer: x = 5 Probelm #2 Directions solve equation ã(4x+1) + 3 = 0 My answer x = 2 Problem #3 Directions solve equation ã(2y+7)+4=y My answer … 
Easy (yet confusing) Algebra
These problems are pretty easy, but confusing (as stated in the subject line). 2x+5x=10 3x+42x=11 I know there are supposed to be two answers for each problem, for the first problem I got 5 and for the second problem I got 7. … 
QUADRATIC EQUATION
r^2+8r=48 First bring the 48 to the other side of the equation r^2+8r48=0 Now you can use the quadratic formula. x=b+sqrt(b^2(4ac)) ¯¯¯¯¯¯¯2a¯¯¯¯¯¯¯¯¯ So... a=1 b=8 c=48 Now just plug the numbers into the quadratic … 
check math
Here are my answers. Can you check if I got the right answers? 
Quadratic Equations
The question is Is the point (3, 2) a solution of the intersection of the following set of quadratic equations: Y < X^2 X^2 + Y^2 < 16 I guess I am somewhat confused by the way it's written. Would I be graphing this to find … 
precalculus
can you check my answers? Find Pk + 1 if Pk = 7 + 13 + 19 + ...+[6(k  1)+1] + (6k + 1) 7 + 13 + 19 + …+[6(k  1) + 1] + (6k + 1) + [6(k + 1) + 1] 8 + 14 + 20 + …+[7(k  1) + 1] + (7k + 1) 7 + 13 + 19 + …+(6k + 1) 7 + 13 + 19 
can you check my answers precalculus
can you check my answers? Find Pk + 1 if Pk = 7 + 13 + 19 + ...+[6(k  1)+1] + (6k + 1) 7 + 13 + 19 + …+[6(k  1) + 1] + (6k + 1) + [6(k + 1) + 1] 8 + 14 + 20 + …+[7(k  1) + 1] + (7k + 1) 7 + 13 + 19 + …+(6k + 1) 7 + 13 + 19 
precalculus
Are my answers correct if not which are right?