Can you please help with Z^4 + Z^2 = 1/4? Answer: Z=+/- 0.455?

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Use the substitution y = z^2 to get
y^2 + y = 1/4
y^2 + y - 1/4 = 0
4y^2 + 4y -1 = 0

Now use the quadratic formula to get y. There will be two values one positive and one negative. You only want the positive one for the next stage.

Then replace y by z^2 and square root....Show more

well if we rearrange then we get:

z^4 + z^2 - 1/4 = 0
or 4z^4 + 4z^2 - 1 = 0

now if we substitute y=z^2 we get:

4y^2 + 4y - 1 = 0

this is a quadratic which we can solve using the formula so:

y = (-4 +- sqrt(16 - 4(4)(-1)))/8 = (-1 +- sqrt(2))/2

now since y=z^2 we have:

z^2 = (-1 +- sqrt(2))/2

If you are doing further maths A-level then you know that there will be complex roots here, but if not dont worry, the real solutions are given when the denominator is positive, since z^2 is positive too so we take the root with the plus sign inbetween:

z = sqrt((-1 + sqrt(2))/2) which works out to be +- 0.455...Show more

Z^4 + Z² = ¼
let Z² = x
=>
x² + x = ¼
x² + x - ¼ = 0
Get roots using the equation: [-b±√(b² - 4ac)] / 2a
x = [-b±√(b² - 4ac)] / 2a
x = [-(1)±√(1² - 4(1)(-¼))] / 2(1)
x = [-1±√2] / 2
x = [-1±1∙4142...] / 2
x = - 2∙414.../2 or 0∙4142.../2
x = -1∙207... or 0∙207...

But x = z²
=> z² = -1∙207... or 0∙207...
z = ±√-1∙207... or ±√0∙207...
z = ± 1∙098..i or ± 0∙455......Show more

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You can substitute z^2 by y, then the equation will be like this:

y^2+y-1/4=0
4y^2+4y-1=0
y=(-4+-sqrt(16-4(4)(-1)))/2(4)
y=(-4+-sqrt(32))/8
y=-1/2+-(1/2)sqrt2
y=0.2071 and y=-1.2071

y=z^2

z^2=0.2071
z=sqrt0.2071
z=+-0.4550

I hope this can useful

David...Show more