Answer:
x=1, -1, 5, and -5.
Explanation:
Given the equation:
[tex]x^4-26x^2=-25[/tex]First, bring the constant term to the left-hand side of the equation:
[tex]x^4-26x^2+25=0[/tex]Next, rewrite the middle term using factors of 25x⁴.
[tex]x^4-25x^2-x^2+25=0[/tex]Factor the first two and last two terms:
[tex]\begin{gathered} x^2(x^2-25)-1(x^2-25)=0 \\ (x^2-1)(x^2-25)=0 \\ (x^2-1)(x^2-5^2)=0 \end{gathered}[/tex]Using the principle of difference of two squares: a²-b²=(a-b)(a+b)
[tex]\begin{gathered} (x-1)(x+1)(x-5)(x+5)=0 \\ \implies x-1=0\text{ or }x+1=0\text{ or }x-5=0\text{ or }x+5=0 \\ \operatorname{\implies}x=1\text{ or }x=-1\text{ or }x=5\text{ or }x=-5 \end{gathered}[/tex]The solutions are 1, -1, 5, and -5.