Find one value of x that is a solution to the equation: {:[(x^(2)-6)^(2)=-3x^(2)+18],[x=◻quad";-x "]:}

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Find one value of  x that is a solution to the equation:  {:[(x^(2)-6)^(2)=-3x^(2)+18],[x=◻quad";-x "]:}

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Analyze the system: Analyze the given system of equations. We have a system of equations: 1. $(x^2 - 6)^2 = -3x^2 + 18$ 2. $x = \square$ or $-x$  We need to find a value of x that satisfies both equations.Solve the first equation: Solve the first equation. $(x^2 - 6)^2 = -3x^2 + 18$  Expand the left side of the equation: $x^4 - 12x^2 + 36 = -3x^2 + 18$  Move all terms to one side to set the equation to zero: $x^4 - 12x^2 + 3x^2 - 18 = 0$  Combine like terms: $x^4 - 9x^2 - 18 = 0$  This is a quadratic equation in terms of $x^2$.Factor the quadratic equation: Factor the quadratic equation. $x^4 - 9x^2 - 18 = 0$  We need to find two numbers that multiply to -18 and add up to -9. The numbers -6 and 3 satisfy these conditions.  So we can write the equation as: $(x^2 - 6)(x^2 + 3) = 0$  Now we have two separate equations: 1. $x^2 - 6 = 0$ 2. $x^2 + 3 = 0$Solve the first separate equation: Solve the first separate equation. $x^2 - 6 = 0$  Add 6 to both sides: $x^2 = 6$  Take the square root of both sides: $x = \sqrt{6}$ or $x = -\sqrt{6}$  These are two possible values for x.Solve the second separate equation: Solve the second separate equation. $x^2 + 3 = 0$  Subtract 3 from both sides: $x^2 = -3$  Since the square of a real number cannot be negative, there are no real solutions to this equation.Check solutions for the second equation: Check which values of x satisfy the second equation of the system. The second equation of the system is $x = \square$ or $-x$, which means we are looking for a value of x that is either positive or negative.  From Step 4, we found two values for x: $\sqrt{6}$ and $-\sqrt{6}$. Both of these values satisfy the second equation since one is positive and the other is negative.  Therefore, we can choose either $\sqrt{6}$ or $-\sqrt{6}$ as the value of x that satisfies the system of equations.

Find one value of $x$ that is a solution to the equation: $\newline$ $\begin{array}{l} \left(x^{2}-6\right)^{2}=-3 x^{2}+18 \\ x=\square \end{array}$

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Find one value of x x x that is a solution to the equation:\newline(x2−6)2=−3x2+18x=□ \begin{array}{l} \left(x^{2}-6\right)^{2}=-3 x^{2}+18 \\ x=\square \end{array} (x2−6)2=−3x2+18x=□​

Find one value of $x$ that is a solution to the equation: $\newline$ $\begin{array}{l} \left(x^{2}-6\right)^{2}=-3 x^{2}+18 \\ x=\square \end{array}$