SOLUTION: Find four roots of the following equation: 2x^4 +5x^2 = 207 * I plugged into my calculator and only got 2 roots, x=-3 and x=3, but how do you get the rest? Thanks :)

Algebra ->  Expressions -> SOLUTION: Find four roots of the following equation: 2x^4 +5x^2 = 207 * I plugged into my calculator and only got 2 roots, x=-3 and x=3, but how do you get the rest? Thanks :)      Log On


   



Question 1013018: Find four roots of the following equation:
2x^4 +5x^2 = 207
* I plugged into my calculator and only got 2 roots, x=-3 and x=3, but how do you get the rest? Thanks :)

Found 4 solutions by ikleyn, stanbon, macston, MathLover1:
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find four roots of the following equation:
2x^4 +5x^2 = 207
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Introduce new variable for your convenience, y = x%5E2.

Then your equation takes the form

2y%5E2+%2B+5y+-+207 = 0.

Apply the quadratic formula. You will get

y%5B1%2C2%5D = %28-5+%2B-+sqrt%2825+%2B4%2A2%2A207%29%29%2F4%29 = %28-5+%2B-+sqrt%281681%29%29%2F4 = %28-5+%2B-+41%29%2F4.

So, y%5B1%5D = 9, y%5B2%5D = -23%2F2.

Now you need to solve this two quadratic equations:

1)  x%5E2 = 9. It has two solutions:  x = 3  and  x = -3.

2)  x%5E2 = -23%2F2. It has two solutions:  x = i%2Asqrt%2823%2F2%29  and  x = -i%2Asqrt%2823%2F2%29.

Thus you got four solutions of your original equation.


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find four roots of the following equation:
2x^4 +5x^2 = 207
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Divide by x^2-9 to get 2x^2 + 23
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Solve 2x^2 + 23 = 0
2x^2 = -23
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x^2 = -11.5
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x = isqrt(11.5) or x = -isqrt(11.5)
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Cheers,
Stan H.
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Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let y=x^2
.
2x^4+5x^2=207 . Replace x^2 with y.
2y^2+5y-207=0
(y-9)(2y+23)=0
y-9=0 . or . 2y+23=0
y=9 . or . 2y=-23
y=9 . or . y=-23/2 . Replace y with x^2
x^2=9 . or . x^2=-23/2
x=+/- 3 . or x=+/-sqrt%2823%2F2%29i=+/-%28sqrt%2846%29%2F2%29i

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E4+%2B5x%5E2+=+207 write in standard form
+2x%5E4%2B5x%5E2-207+=0
Factor:
+2x%5E4%2B5x%5E2-207+=0
The first term is, 2x%5E4 its coefficient is 2 .
The middle term is, 5x%5E2+ its coefficient is 5 .
The last term, "the constant", is -207+
Multiply the coefficient of the first term by the constant +2+%28-207%29+=+-414+
Find two factors of -414 whose sum equals the coefficient of the middle term, which is +5: these are -18 and 23
Rewrite the polynomial splitting the middle term using the two factors found above, -18 and 23:

+2x%5E4+-+18x%5E2+%2B+23x%5E2+-+207=0 ............group first two terms together and second two terms together
%282x%5E4+-18x%5E2%29+%2B+%2823x%5E2+-207%29=0
+2x%5E2%28x%5E2+-+9%29+%2B+23%28x%5E2+-9%29=0
%282x%5E2%2B+23%29%28x%5E2+-+9%29=0
solutions:
if 2x%5E2%2B+23=0 =>2x%5E2=-+23 =>x%5E2=-+23%2F2 =>x=sqrt%28-+23%2F2%29=>x=++i%2Asqrt%28+23%2F2%29%29
two complex solutions are: x=i%2Asqrt%2823%2F2%29 and +x=-i%2Asqrt%2823%2F2%29+
if %28x%5E2+-+9%29=0 =>x%5E2+=9 =>x=sqrt%289%29 =>
two real solutions: x=3 and x=-3