What are the real zeros of the polynomial P(x)=x4−4x2−1? Round to the nearest tenth.

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BlackTom AI · Accepted Answer

We start by restating the problem in our own words to focus on what we need to find: the real zeros of P(x) = x^4 − 4x^2 − 1, rounded to the nearest tenth.
First, let y = x^2. The polynomial becomes a quadratic in y: y^2 − 4y − 1 = 0.
Solve this quadratic for y using the quadratic formula: y = [4 ± sqrt( (-4)^2 − 4·1·(-1) )] / (2·1) = [4 ± sqrt(16 + 4)] / 2 = [4 ± sqrt(20)] / 2 = [4 ± 2√5] / 2 = 2 ± √5.
Compute the numerical values: √5 ≈ 2.236, so y1 = 2 + 2.236 ≈ 4.236 and y2 = 2 − 2.236 ≈ −0.236.
Since y = x^2 cannot be negative, we discard y2 and keep y = x^2 ≈ 4.236.
Now solve for x: x^2 = 4.236, so x = ±√4.236. The square root of 4.236 is approximately 2.059.
Rounding to the nearest tenth gives x ≈ ±2.1, so the real zeros are approximately −2.1 and 2.1.
Note: If you had to choose from multiple options, you would select the pair that lists −2.1 and 2.1, reflecting both positive and negative square roots of the positive y-value.

ALGII141-027 4.6 Practice Problems

What are the real zeros of the polynomial P(x)=x4−4x2−1? Round to the nearest tenth.

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