SOLUTION: identify the root to the nearest tenth. y=2x^4+x^2-3

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Question 7077: identify the root to the nearest tenth.
y=2x^4+x^2-3
Found 2 solutions by ichudov, khwang:
Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
You need to solve equation
2x%5E4%2Bx%5E2-3+=+0
Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: x=1,x=-1.
  • Graphical form: Equation 2x%5E4%2Bx%5E2-3=0 was fully solved.
  • Text form: 2x^4+x^2-3=0 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+2x%5E4%2Bx%5E2-3=0+%29
    For tutors: simplify_cartoon( 2x^4+x^2-3=0 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at highlight_red%28+2%2Ax%5E4%2Bx%5E2-3+%29=0.
Equation highlight_red%28+2%2Ax%5E4%2Bx%5E2-3=0+%29 is a quadratic equation, if we substitute Z=x%5E2: and has solutions Z= 1 and solution x=root%28+2%2C+Z+%29: 1,-1
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: x=1,x=-1.

Universal Simplifier and Solver


Done!

Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
2x^4+x^2-3=0
-> (2x^2 + 3) (x^2 -1) = 0
--> (2x^2 + 3) (x -1)(x+1) = 0
--> x =1 ,-1 or +i sqrt(3/2) or -i sqrt(3/2) (or +/-sqrt(6)/2)

I don't do the approx. values that are left for you.
(two real roots and two complex conjugates).
Kenny