Answer :
Let's solve each of these equations step-by-step:
Equation 22: [tex]\( x^4 + 6x^2 + 5 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( u^2 + 6u + 5 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( u^2 + 6u + 5 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = -5 \)[/tex] and [tex]\( u = -1 \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = -5 \)[/tex], [tex]\( x^2 = -5 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{5}i \)[/tex].
- For [tex]\( u = -1 \)[/tex], [tex]\( x^2 = -1 \)[/tex] gives solutions [tex]\( x = \pm i \)[/tex].
Equation 23: [tex]\( x^4 - 3x^2 - 10 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( u^2 - 3u - 10 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( u^2 - 3u - 10 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = 5 \)[/tex] and [tex]\( u = -2 \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = 5 \)[/tex], [tex]\( x^2 = 5 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{5} \)[/tex].
- For [tex]\( u = -2 \)[/tex], [tex]\( x^2 = -2 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{2}i \)[/tex].
Equation 24: [tex]\( 4x^4 - 14x^2 + 12 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( 4u^2 - 14u + 12 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( 4u^2 - 14u + 12 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = \frac{6}{4} \)[/tex] and [tex]\( u = 2 \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = \frac{6}{4} = \frac{3}{2} \)[/tex], [tex]\( x^2 = \frac{3}{2} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{6}}{2} \)[/tex].
- For [tex]\( u = 2 \)[/tex], [tex]\( x^2 = 2 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{2} \)[/tex].
Equation 25: [tex]\( 9x^4 - 27x^2 + 20 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( 9u^2 - 27u + 20 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( 9u^2 - 27u + 20 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = \frac{15}{9} \)[/tex] and [tex]\( u = \frac{12}{9} \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = \frac{15}{9} = \frac{5}{3} \)[/tex], [tex]\( x^2 = \frac{5}{3} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{15}}{3} \)[/tex].
- For [tex]\( u = \frac{12}{9} = \frac{4}{3} \)[/tex], [tex]\( x^2 = \frac{4}{3} \)[/tex] gives solutions [tex]\( x = \pm \frac{2\sqrt{3}}{3} \)[/tex].
Equation 26: [tex]\( 4x^4 - 5x^2 - 6 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( 4u^2 - 5u - 6 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( 4u^2 - 5u - 6 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = \frac{3}{4} \)[/tex] and [tex]\( u = -2 \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = \frac{3}{4} \)[/tex], [tex]\( x^2 = \frac{3}{4} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{3}}{2}i \)[/tex].
- For [tex]\( u = -2 \)[/tex], [tex]\( x^2 = -2 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{2} \)[/tex].
Equation 27: [tex]\( 24x^4 + 14x^2 - 3 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( 24u^2 + 14u - 3 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( 24u^2 + 14u - 3 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = \frac{3}{4} \)[/tex] and [tex]\( u = -\frac{1}{6} \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = \frac{3}{4} \)[/tex], [tex]\( x^2 = \frac{3}{4} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{3}}{2}i \)[/tex].
- For [tex]\( u = -\frac{1}{6} \)[/tex], [tex]\( x^2 = -\frac{1}{6} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{6}}{6} \)[/tex].
These steps lead us to the particular solutions of each equation mentioned above.
Equation 22: [tex]\( x^4 + 6x^2 + 5 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( u^2 + 6u + 5 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( u^2 + 6u + 5 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = -5 \)[/tex] and [tex]\( u = -1 \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = -5 \)[/tex], [tex]\( x^2 = -5 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{5}i \)[/tex].
- For [tex]\( u = -1 \)[/tex], [tex]\( x^2 = -1 \)[/tex] gives solutions [tex]\( x = \pm i \)[/tex].
Equation 23: [tex]\( x^4 - 3x^2 - 10 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( u^2 - 3u - 10 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( u^2 - 3u - 10 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = 5 \)[/tex] and [tex]\( u = -2 \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = 5 \)[/tex], [tex]\( x^2 = 5 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{5} \)[/tex].
- For [tex]\( u = -2 \)[/tex], [tex]\( x^2 = -2 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{2}i \)[/tex].
Equation 24: [tex]\( 4x^4 - 14x^2 + 12 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( 4u^2 - 14u + 12 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( 4u^2 - 14u + 12 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = \frac{6}{4} \)[/tex] and [tex]\( u = 2 \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = \frac{6}{4} = \frac{3}{2} \)[/tex], [tex]\( x^2 = \frac{3}{2} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{6}}{2} \)[/tex].
- For [tex]\( u = 2 \)[/tex], [tex]\( x^2 = 2 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{2} \)[/tex].
Equation 25: [tex]\( 9x^4 - 27x^2 + 20 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( 9u^2 - 27u + 20 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( 9u^2 - 27u + 20 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = \frac{15}{9} \)[/tex] and [tex]\( u = \frac{12}{9} \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = \frac{15}{9} = \frac{5}{3} \)[/tex], [tex]\( x^2 = \frac{5}{3} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{15}}{3} \)[/tex].
- For [tex]\( u = \frac{12}{9} = \frac{4}{3} \)[/tex], [tex]\( x^2 = \frac{4}{3} \)[/tex] gives solutions [tex]\( x = \pm \frac{2\sqrt{3}}{3} \)[/tex].
Equation 26: [tex]\( 4x^4 - 5x^2 - 6 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( 4u^2 - 5u - 6 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( 4u^2 - 5u - 6 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = \frac{3}{4} \)[/tex] and [tex]\( u = -2 \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = \frac{3}{4} \)[/tex], [tex]\( x^2 = \frac{3}{4} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{3}}{2}i \)[/tex].
- For [tex]\( u = -2 \)[/tex], [tex]\( x^2 = -2 \)[/tex] gives solutions [tex]\( x = \pm \sqrt{2} \)[/tex].
Equation 27: [tex]\( 24x^4 + 14x^2 - 3 = 0 \)[/tex]
1. Substitute [tex]\( u = x^2 \)[/tex], so the equation becomes [tex]\( 24u^2 + 14u - 3 = 0 \)[/tex].
2. Solve the quadratic equation: [tex]\( 24u^2 + 14u - 3 = 0 \)[/tex].
3. The solutions for [tex]\( u \)[/tex] are [tex]\( u = \frac{3}{4} \)[/tex] and [tex]\( u = -\frac{1}{6} \)[/tex].
4. Substitute back [tex]\( x^2 = u \)[/tex]:
- For [tex]\( u = \frac{3}{4} \)[/tex], [tex]\( x^2 = \frac{3}{4} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{3}}{2}i \)[/tex].
- For [tex]\( u = -\frac{1}{6} \)[/tex], [tex]\( x^2 = -\frac{1}{6} \)[/tex] gives solutions [tex]\( x = \pm \frac{\sqrt{6}}{6} \)[/tex].
These steps lead us to the particular solutions of each equation mentioned above.
