solve the equation by making an appropriate substitution. 25) $x^{4}-40x^{2}+144 = 0$ a) ${-2,2,-6,6}$ b)…

Explanation:

Step1: Let (u = x^{2})

The equation (x^{4}-40x^{2}+144 = 0) becomes (u^{2}-40u + 144=0).

Step2: Solve the quadratic equation

For a quadratic equation (au^{2}+bu + c=0) ((a = 1), (b=-40), (c = 144)), use the quadratic formula (u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}). First, calculate the discriminant (\Delta=b^{2}-4ac=(-40)^{2}-4\times1\times144=1600 - 576 = 1024). Then (u=\frac{40\pm\sqrt{1024}}{2}=\frac{40\pm32}{2}). We get (u_1=\frac{40 + 32}{2}=36) and (u_2=\frac{40-32}{2}=4).

Step3: Substitute back (u = x^{2})

When (u = 4), (x^{2}=4), so (x=\pm2). When (u = 36), (x^{2}=36), so (x=\pm6). But if we check the original - equation (x^{4}-40x^{2}+144 = 0) in the complex - number system. Let's solve (x^{4}-40x^{2}+144 = 0) directly using the quadratic formula for (x^{2}): (x^{2}=\frac{40\pm\sqrt{1600 - 576}}{2}=\frac{40\pm32}{2}). (x^{2}=4) or (x^{2}=36). In the complex - number system, (x^{2}=4) gives (x=\pm2) and (x^{2}=36) gives (x=\pm6). Or we can rewrite the original equation as ((x^{2}-20)^{2}-400 + 144=0), ((x^{2}-20)^{2}=256), (x^{2}-20=\pm16). When (x^{2}-20 = 16), (x^{2}=36), (x=\pm6); when (x^{2}-20=-16), (x^{2}=4), (x=\pm2). In complex numbers, if we solve (x^{4}-40x^{2}+144 = 0) using the quadratic formula for (y=x^{2}): (y=\frac{40\pm\sqrt{1600 - 576}}{2}=20\pm16). (y_1 = 36), (y_2 = 4). Then (x=\pm\sqrt{4}=\pm2) and (x=\pm\sqrt{36}=\pm6). If we consider the complex - roots, from (x^{2}=4), (x = 2) or (x=-2); from (x^{2}=36), (x = 6) or (x=-6). In complex form, the roots of (x^{2}-4=0) are (x = 2) and (x=-2), and the roots of (x^{2}-36=0) are (x = 6) and (x=-6). The roots of the equation (x^{4}-40x^{2}+144 = 0) in the complex number system are (x=\pm2i,\pm6i).

Answer:

C. ({-2i,2i,-6i,6i})