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

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.

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

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...

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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

z^4 +z^2=1/4

=>4z^4 +4z^2=1

=>(2z^2)^2 +2.2z^2 .1 +(1)^2= 1+1

=>(2z^2 +1)^2 = 2

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

=>2z^2 +1=+/- 1.414

=>2z^2 = +/- 1.414 - 1

when, 2z^2 =1.414 - 1

=>2z^2 =0.414

=>z^2=0.207

=>z=+/- 0.455

and when, 2z^2= -1.414 -1

=>2z^2= -2.414

=>z^2= - 1.207

==>[not possible coz square of a number can't be negetive]

[when imaginary roots are not required

z^2= 1.207.i^2

=>z=+/- 1.099.i

where i^2= - 1]

So, z= +/- 0.455<==ANSWER

Not sure where you got those answers, but here's how I got mine.

z^4 + z^2 = 1/4

z^2(z^2 + 1) = 1/4

so you get z^2 = 1/4 and (z^2 + 1) = 1/4

solving for z in z^2 = 1/4 you get z = 1/2 and z = -1/2

solving for (z^2 + 1) = 1/4 you get z = 3/4i and -3/4i.

put y=Z². Then we ave

y² + y - 0.25 = 0

discriminant : 1 - (-0.25*4) = 2

=> y = (-1 +- sqrt(2))/2 = Z²

so Z = +- sqrt(-0.5 +- 0.5 sqrt(2))

= +- 0.4550899 and +- 1.0986841 i (complex roots)

z^4+z^2=1/4

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

(z^2)^2+z^2-(1/2)^2=0

(z^2)^2+(2*1/2*z^2)-(1/2)^2=0

(z^2-1/2)^2=0

z^2-1/2=0

z^2=+0.5

z=+/-0.455