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# How do I find the zeroes of this polynomial function?

p(x)=x^4-x^2+0.25

can you show me the steps?

### 13 Answers

- lenpol7Lv 72 weeks ago
Let x^2 - y

Substitute

p(y) = y^2 - y + 0.5^2 = 0

Complete the Square

(y - 0.5)^2 - (0.5)^2 = - (0.5)^2

(y - 0.5)^2 = -0.25 +0.25 = 0

( y -0.5)^2 = 0

y - 0.5 = 0

Back substitute

x^2 - 0.5 = 0

x^2 = 0.5

x = +/-sqrt(0.5)

x = +/-0.7071... Which agrees with the graph.

- charlatanLv 72 weeks ago
i was really stumped by the way the question was framed.

as i know one equates a quadratic equation to "0",

and tries to figure out roots.

- RaymondLv 72 weeks ago
It is a quadratic if you take your variable to be "x^2"

p(x) = (x^2)^2 - (x^2) + 0.25

use a=1, b= -1, c = 0.25

(Remember, our variable is x^2, not x, for this to be a quadratic)

(x^2) = [ -b ± √(b^2 - 4ac) ] / 2a

which becomes

(x^2) = [ 1 ± √(1 - 1) ] / 2

(x^2) = [ 1 ± √(0) ] / 2

x^2 = 1/2

Therefore

x = ± √(1/2)

are the zeros (or the roots - in this context, it means the same thing)

- TomVLv 72 weeks ago
p(x) is quadratic in u if u = x². Solve the quadratic.

u = x²

p(u) = u² - u + 1/4 = 0

u² - u + 1/4 = 0

u = 1/2

x = ±√(1/2) = ±√2/2

Or, plot the polynomial and observe the intersections with the x-axis, x = ±0.707

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- ?Lv 72 weeks ago
Solving x^4 - x^2 + 0.25 = 0

Let u = x^2

u^2 - u + 0.25 = 0

Apply quadratic formula.

u = 1/2

Convert back to x.

x^2 = 1/2

x = ±1/√2

- llafferLv 72 weeks ago
p(x) = x⁴ - x² + 0.25

Find the zeros, so:

0 = x⁴ - x² + 0.25

This is a perfect square trinomial, so:

0 = (x² - 0.5)²

Square root of both sides:

0 = x² - 0.5

x² = 0.5

Square root of both sides again:

x = ± √(0.5)

Let's simplify that a little. If we turn that into a fraction we can then rationalize the denominator:

x = ± √(1/2)

x = ± 1/√2

x = ± √(2)/2

or:

x = ±(1/2)√2

- KrishnamurthyLv 72 weeks ago
p(x) = x^4 - x^2 + 0.25

p(x) = (x - 0.707107)^2 (x + 0.707107)^2

Roots:

x ≈ -0.707107

x ≈ 0.707107

- billrussell42Lv 72 weeks ago
x⁴ – x² + 1/4 = 0

substitute z = x²

z² – z + 1/4 = 0

quadratic equation:to solve ax² + bx + c = 0x = [–b ± √(b²–4ac)] / 2az = [1 ± √(1–1] / 2

z = 1/2

z = x²

x = ±√(1/2)

check

x⁴ – x² + 1/4 = 0

x = √(1/2)

(√(1/2))⁴ – (√(1/2))² + 1/4 = 0

1/4 – 1/2 + 1/4 = 0

ok

x = –√(1/2)

(–√(1/2))⁴ – (–√(1/2))² + 1/4 = 0

1/4 – 1/2 + 1/4 = 0

ok