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Find one value of x that is a solution to the equation: {:[(x^(2)+1)^(2)-5x^(2)-5=0],[x=◻]:}

Find one value of x x that is a solution to the equation:\newline(x2+1)2−5x2−5=0x=□ \begin{array}{l} \left(x^{2}+1\right)^{2}-5 x^{2}-5=0 \\ x=\square \end{array}

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Full solution

Q. Find one value of x x that is a solution to the equation:\newline(x2+1)2−5x2−5=0x=□ \begin{array}{l} \left(x^{2}+1\right)^{2}-5 x^{2}-5=0 \\ x=\square \end{array}
  1. Expand the equation: Let's start by expanding the equation (x2+1)2−5x2−5=0(x^2 + 1)^2 - 5x^2 - 5 = 0.\newlineWe will expand (x2+1)2(x^2 + 1)^2 to x4+2x2+1x^4 + 2x^2 + 1.\newlineSo the equation becomes x4+2x2+1−5x2−5=0x^4 + 2x^2 + 1 - 5x^2 - 5 = 0.
  2. Combine like terms: Now, let's combine like terms. x4+2x2−5x2+1−5=0x^4 + 2x^2 - 5x^2 + 1 - 5 = 0 simplifies to x4−3x2−4=0x^4 - 3x^2 - 4 = 0.
  3. Factor the equation: We can factor this equation as a quadratic in terms of x2x^2.\newlineThis gives us (x2−4)(x2+1)=0(x^2 - 4)(x^2 + 1) = 0.
  4. Solve x2−4=0x^2 - 4 = 0: Now we have two factors set to zero: x2−4=0x^2 - 4 = 0 and x2+1=0x^2 + 1 = 0.\newlineWe can solve each of these separately.
  5. Solve x2+1=0x^2 + 1 = 0: First, let's solve x2−4=0x^2 - 4 = 0.\newlineAdding 44 to both sides gives us x2=4x^2 = 4.\newlineTaking the square root of both sides gives us x=±2x = \pm 2.
  6. Solve x2+1=0x^2 + 1 = 0: First, let's solve x2−4=0x^2 - 4 = 0.\newlineAdding 44 to both sides gives us x2=4x^2 = 4.\newlineTaking the square root of both sides gives us x=±2x = \pm2.Now let's solve x2+1=0x^2 + 1 = 0.\newlineSubtracting 11 from both sides gives us x2=−1x^2 = -1.\newlineHowever, since x2=−1x^2 = -1 has no real solution (as the square of a real number cannot be negative), we will not consider this further.

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