Question

solve.
$x^{4}-24x^{2}+144 = 0$
write your answer as a list of values separated by commas. simplify all radicals.

Explanation:

Step1: Let \( y = x^2 \), rewrite the equation.

The equation \( x^4 - 24x^2 + 144 = 0 \) can be rewritten as \( y^2 - 24y + 144 = 0 \) by substituting \( y = x^2 \).

Step2: Solve the quadratic equation for \( y \).

We recognize that \( y^2 - 24y + 144 \) is a perfect square trinomial, since \( (-12)^2=144 \) and \( 2\times(-12)\times y=-24y \). So, \( y^2 - 24y + 144=(y - 12)^2 \).
Setting \( (y - 12)^2 = 0 \), we get \( y - 12 = 0 \), so \( y = 12 \).

Step3: Substitute back \( y = x^2 \) and solve for \( x \).

Since \( y = x^2 = 12 \), we solve for \( x \) by taking the square root of both sides. So, \( x=\pm\sqrt{12} \). Simplifying \( \sqrt{12} \), we know that \( 12 = 4\times3 \), so \( \sqrt{12}=\sqrt{4\times3}=2\sqrt{3} \). Thus, \( x = 2\sqrt{3}, - 2\sqrt{3} \). But we should also check if there are repeated roots. Since the original equation is a quartic (degree 4) and we had a repeated root for \( y \) (double root at \( y = 12 \)), when we solve \( x^2=12 \), each solution for \( x \) is also a double root? Wait, no. Wait, the quadratic in \( y \) was \( (y - 12)^2=0 \), so \( y = 12 \) is a double root. Then \( x^2=12 \) gives \( x=\pm\sqrt{12} \), and each of these is a root of multiplicity 2? Wait, let's check the original equation. Let's plug \( x = 2\sqrt{3} \) into \( x^4-24x^2 + 144 \). \( (2\sqrt{3})^4-24\times(2\sqrt{3})^2+144=(16\times9)-24\times(4\times3)+144 = 144-288 + 144=0 \). Similarly for \( x=-2\sqrt{3} \). But actually, since the equation in \( y \) had a double root, the original quartic equation has roots \( x = 2\sqrt{3},2\sqrt{3},-2\sqrt{3},-2\sqrt{3} \)? Wait, no, wait. Wait, the substitution \( y=x^2 \), so when we have \( (y - 12)^2=0 \), that means \( y = 12 \) is a root of multiplicity 2. Then \( x^2=12 \) is equivalent to \( x^2-12 = 0 \), which factors as \( (x - \sqrt{12})(x+\sqrt{12})=0 \), but since \( y - 12=(x^2 - 12)=(x - \sqrt{12})(x+\sqrt{12}) \), so \( (y - 12)^2=(x^2 - 12)^2=(x - \sqrt{12})^2(x+\sqrt{12})^2 \). So the roots of the original equation are \( x = \sqrt{12}, \sqrt{12}, -\sqrt{12}, -\sqrt{12} \), which simplifies to \( x = 2\sqrt{3},2\sqrt{3},-2\sqrt{3},-2\sqrt{3} \). But when we write the solution as a list of values (with duplicates or not? The problem says "write your answer as a list of values separated by commas". It doesn't specify to remove duplicates, but usually, for polynomial equations, we can list the roots with their multiplicities or just the distinct roots? Wait, let's check the problem statement: "Write your answer as a list of values separated by commas. Simplify all radicals." So probably, we can list the distinct roots, but actually, in the equation \( x^4-24x^2 + 144 = 0 \), the roots are \( 2\sqrt{3},2\sqrt{3},-2\sqrt{3},-2\sqrt{3} \), but when we solve \( x^2 = 12 \), we get \( x=\pm2\sqrt{3} \), and since the original equation is a perfect square of a quadratic, \( (x^2 - 12)^2=0 \), so the roots are \( x = 2\sqrt{3} \) (multiplicity 2) and \( x=-2\sqrt{3} \) (multiplicity 2). But the problem says "write your answer as a list of values separated by commas". So maybe we can write the distinct roots, but let's see: if we consider that when we solve \( x^2=12 \), we have two solutions, each with multiplicity 2, but the problem might just want the distinct solutions. Wait, let's re - examine the steps.

Wait, step 2: the quadratic in \( y \) is \( y^2-24y + 144=(y - 12)^2 \), so \( y = 12 \) (double root). Then \( x^2=12 \), so \( x=\pm\sqrt{12}=\pm2\sqrt{3} \). So the solutions for \( x \) are \( 2\sqrt{3}, - 2\sqrt{3} \), but each of these is a root of multiplicity 2 in the original quartic equation. But the problem says "write your answer as a list of values separated by commas". So we can write \( 2\sqrt{3}, - 2\sqrt{3} \), but actually, since the equation is \( (x^2 - 12)^2=0 \), the roots are \( x = 2\sqrt{3},2\sqrt{3},-2\sqrt{3},-2\sqrt{3} \). But maybe the problem expects the distinct roots, or maybe it's okay to list them with multiplicity? Wait, let's check with the substitution. If we let \( x^2 = 12 \), then \( x=\pm2\sqrt{3} \), and these are the only real roots (since the discriminant of \( x^2 - 12=0 \) is positive, so real roots). So the solutions are \( 2\sqrt{3}, - 2\sqrt{3} \) (and each is a double root, but when writing the list, maybe we can write them as \( 2\sqrt{3},2\sqrt{3},-2\sqrt{3},-2\sqrt{3} \), but that seems redundant. Wait, no, let's solve \( x^4-24x^2 + 144 = 0 \). Let's factor it as \( (x^2 - 12)^2=0 \), so \( x^2=12 \), so \( x=\pm\sqrt{12}=\pm2\sqrt{3} \). So the solutions are \( x = 2\sqrt{3} \) (with multiplicity 2) and \( x=-2\sqrt{3} \) (with multiplicity 2). But the problem says "write your answer as a list of values separated by commas". So maybe we can write \( 2\sqrt{3}, - 2\sqrt{3} \), but actually, in the context of solving equations, sometimes repeated roots are listed, but sometimes not. Wait, let's check with an example. If we have \( (x - 2)^2=0 \), the solution is \( x = 2,2 \), but sometimes it's written as \( x = 2 \) (with multiplicity 2). But the problem here has a quartic, and when we solve \( x^2=12 \), we get two solutions, each with multiplicity 2. But the problem says "simplify all radicals" and "write your answer as a list of values separated by commas". So probably, we can write \( 2\sqrt{3}, - 2\sqrt{3} \), but actually, since the original equation is a perfect square of a quadratic, the roots are \( 2\sqrt{3},2\sqrt{3},-2\sqrt{3},-2\sqrt{3} \). Wait, let's compute \( (x^2 - 12)^2=x^4-24x^2 + 144 \), which is the given equation. So when we set \( (x^2 - 12)^2=0 \), we have \( x^2 - 12 = 0 \) (double root), so \( x^2=12 \), so \( x=\pm\sqrt{12}=\pm2\sqrt{3} \), and each of these is a root of multiplicity 2. But the problem might just want the distinct roots. Let's see, if we plug \( x = 2\sqrt{3} \) into the equation: \( (2\sqrt{3})^4-24\times(2\sqrt{3})^2 + 144=(16\times9)-24\times(4\times3)+144=144 - 288+144 = 0 \). Similarly for \( x=-2\sqrt{3} \). So the solutions are \( 2\sqrt{3}, - 2\sqrt{3} \) (with multiplicity 2 each). But the problem says "write your answer as a list of values separated by commas". So maybe we can write \( 2\sqrt{3}, - 2\sqrt{3} \), but actually, in the equation \( x^4-24x^2 + 144 = 0 \), the roots are \( x = 2\sqrt{3},2\sqrt{3},-2\sqrt{3},-2\sqrt{3} \). However, when we solve \( x^2=12 \), we get two solutions, and since the original equation is a quartic, each of these solutions is a double root. But the problem might accept the distinct solutions. Let's confirm with the quadratic substitution. We had \( y = 12 \) (double root), then \( x^2=12 \), so \( x=\pm\sqrt{12}=\pm2\sqrt{3} \). So the solutions are \( 2\sqrt{3}, - 2\sqrt{3} \).

Answer:

\( 2\sqrt{3}, - 2\sqrt{3} \)