Answer :
Let's solve each of the given equations step-by-step:
22. [tex]\( x^4 + 6x^2 + 5 = 0 \)[/tex]
This can be treated as a quadratic equation by substituting [tex]\( y = x^2 \)[/tex]. So, we rewrite it as [tex]\( y^2 + 6y + 5 = 0 \)[/tex].
Now, solve the quadratic equation:
1. Factor the equation: [tex]\( (y + 5)(y + 1) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( y + 5 = 0 \)[/tex] or [tex]\( y + 1 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = -5 \)[/tex] or [tex]\( y = -1 \)[/tex].
Since [tex]\( y = x^2 \)[/tex], we have:
- [tex]\( x^2 = -5 \)[/tex] leads to [tex]\( x = \pm \sqrt{-5} = \pm \sqrt{5}i \)[/tex].
- [tex]\( x^2 = -1 \)[/tex] leads to [tex]\( x = \pm \sqrt{-1} = \pm i \)[/tex].
23. [tex]\( x^4 - 3x^2 - 10 = 0 \)[/tex]
Let [tex]\( y = x^2 \)[/tex], transforming the equation into [tex]\( y^2 - 3y - 10 = 0 \)[/tex].
1. Factor the equation: [tex]\( (y - 5)(y + 2) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( y - 5 = 0 \)[/tex] or [tex]\( y + 2 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = 5 \)[/tex] or [tex]\( y = -2 \)[/tex].
Because [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = 5 \)[/tex] gives [tex]\( x = \pm \sqrt{5} \)[/tex].
- [tex]\( x^2 = -2 \)[/tex] gives [tex]\( x = \pm \sqrt{2}i \)[/tex].
24. [tex]\( 4x^4 - 14x^2 + 12 = 0 \)[/tex]
Substitute [tex]\( y = x^2 \)[/tex], making it [tex]\( 4y^2 - 14y + 12 = 0 \)[/tex].
1. Factor the equation: [tex]\( (2y - 6)(2y - 2) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( 2y - 6 = 0 \)[/tex] or [tex]\( 2y - 2 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = 3 \)[/tex] or [tex]\( y = 1 \)[/tex].
Since [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = 3 \)[/tex] gives [tex]\( x = \pm \sqrt{3} \)[/tex].
- [tex]\( x^2 = 1 \)[/tex] gives [tex]\( x = \pm 1 \)[/tex].
25. [tex]\( 9x^4 - 27x^2 + 20 = 0 \)[/tex]
Let [tex]\( y = x^2 \)[/tex] so that we rewrite it as [tex]\( 9y^2 - 27y + 20 = 0 \)[/tex].
1. Factor the equation: [tex]\( (3y - 5)(3y - 4) = 0 \)[/tex].
2. Set each equation to zero: [tex]\( 3y - 5 = 0 \)[/tex] or [tex]\( 3y - 4 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = \frac{5}{3} \)[/tex] or [tex]\( y = \frac{4}{3} \)[/tex].
Since [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = \frac{5}{3} \)[/tex] gives [tex]\( x = \pm \frac{\sqrt{15}}{3} \)[/tex].
- [tex]\( x^2 = \frac{4}{3} \)[/tex] gives [tex]\( x = \pm \frac{2\sqrt{3}}{3} \)[/tex].
26. [tex]\( 4x^4 - 5x^2 - 6 = 0 \)[/tex]
By setting [tex]\( y = x^2 \)[/tex], we have [tex]\( 4y^2 - 5y - 6 = 0 \)[/tex].
1. Factor the equation: [tex]\( (4y + 3)(y - 2) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( 4y + 3 = 0 \)[/tex] or [tex]\( y - 2 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = -\frac{3}{4} \)[/tex] or [tex]\( y = 2 \)[/tex].
Since [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = -\frac{3}{4} \)[/tex] gives [tex]\( x = \pm \frac{\sqrt{3}}{2} i \)[/tex].
- [tex]\( x^2 = 2 \)[/tex] gives [tex]\( x = \pm \sqrt{2} \)[/tex].
27. [tex]\( 24x^4 + 14x^2 - 3 = 0 \)[/tex]
With [tex]\( y = x^2 \)[/tex], rewrite the equation as [tex]\( 24y^2 + 14y - 3 = 0 \)[/tex].
1. Factor the equation: [tex]\( (6y + 3)(4y - 1) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( 6y + 3 = 0 \)[/tex] or [tex]\( 4y - 1 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = -\frac{1}{2} \)[/tex] or [tex]\( y = \frac{1}{4} \)[/tex].
Since [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = -\frac{1}{2} \)[/tex] gives [tex]\( x = \pm \frac{\sqrt{2}}{2} i \)[/tex].
- [tex]\( x^2 = \frac{1}{4} \)[/tex] gives [tex]\( x = \pm \frac{\sqrt{3}}{6} \)[/tex].
These detailed steps explain the solution for each equation.
22. [tex]\( x^4 + 6x^2 + 5 = 0 \)[/tex]
This can be treated as a quadratic equation by substituting [tex]\( y = x^2 \)[/tex]. So, we rewrite it as [tex]\( y^2 + 6y + 5 = 0 \)[/tex].
Now, solve the quadratic equation:
1. Factor the equation: [tex]\( (y + 5)(y + 1) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( y + 5 = 0 \)[/tex] or [tex]\( y + 1 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = -5 \)[/tex] or [tex]\( y = -1 \)[/tex].
Since [tex]\( y = x^2 \)[/tex], we have:
- [tex]\( x^2 = -5 \)[/tex] leads to [tex]\( x = \pm \sqrt{-5} = \pm \sqrt{5}i \)[/tex].
- [tex]\( x^2 = -1 \)[/tex] leads to [tex]\( x = \pm \sqrt{-1} = \pm i \)[/tex].
23. [tex]\( x^4 - 3x^2 - 10 = 0 \)[/tex]
Let [tex]\( y = x^2 \)[/tex], transforming the equation into [tex]\( y^2 - 3y - 10 = 0 \)[/tex].
1. Factor the equation: [tex]\( (y - 5)(y + 2) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( y - 5 = 0 \)[/tex] or [tex]\( y + 2 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = 5 \)[/tex] or [tex]\( y = -2 \)[/tex].
Because [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = 5 \)[/tex] gives [tex]\( x = \pm \sqrt{5} \)[/tex].
- [tex]\( x^2 = -2 \)[/tex] gives [tex]\( x = \pm \sqrt{2}i \)[/tex].
24. [tex]\( 4x^4 - 14x^2 + 12 = 0 \)[/tex]
Substitute [tex]\( y = x^2 \)[/tex], making it [tex]\( 4y^2 - 14y + 12 = 0 \)[/tex].
1. Factor the equation: [tex]\( (2y - 6)(2y - 2) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( 2y - 6 = 0 \)[/tex] or [tex]\( 2y - 2 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = 3 \)[/tex] or [tex]\( y = 1 \)[/tex].
Since [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = 3 \)[/tex] gives [tex]\( x = \pm \sqrt{3} \)[/tex].
- [tex]\( x^2 = 1 \)[/tex] gives [tex]\( x = \pm 1 \)[/tex].
25. [tex]\( 9x^4 - 27x^2 + 20 = 0 \)[/tex]
Let [tex]\( y = x^2 \)[/tex] so that we rewrite it as [tex]\( 9y^2 - 27y + 20 = 0 \)[/tex].
1. Factor the equation: [tex]\( (3y - 5)(3y - 4) = 0 \)[/tex].
2. Set each equation to zero: [tex]\( 3y - 5 = 0 \)[/tex] or [tex]\( 3y - 4 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = \frac{5}{3} \)[/tex] or [tex]\( y = \frac{4}{3} \)[/tex].
Since [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = \frac{5}{3} \)[/tex] gives [tex]\( x = \pm \frac{\sqrt{15}}{3} \)[/tex].
- [tex]\( x^2 = \frac{4}{3} \)[/tex] gives [tex]\( x = \pm \frac{2\sqrt{3}}{3} \)[/tex].
26. [tex]\( 4x^4 - 5x^2 - 6 = 0 \)[/tex]
By setting [tex]\( y = x^2 \)[/tex], we have [tex]\( 4y^2 - 5y - 6 = 0 \)[/tex].
1. Factor the equation: [tex]\( (4y + 3)(y - 2) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( 4y + 3 = 0 \)[/tex] or [tex]\( y - 2 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = -\frac{3}{4} \)[/tex] or [tex]\( y = 2 \)[/tex].
Since [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = -\frac{3}{4} \)[/tex] gives [tex]\( x = \pm \frac{\sqrt{3}}{2} i \)[/tex].
- [tex]\( x^2 = 2 \)[/tex] gives [tex]\( x = \pm \sqrt{2} \)[/tex].
27. [tex]\( 24x^4 + 14x^2 - 3 = 0 \)[/tex]
With [tex]\( y = x^2 \)[/tex], rewrite the equation as [tex]\( 24y^2 + 14y - 3 = 0 \)[/tex].
1. Factor the equation: [tex]\( (6y + 3)(4y - 1) = 0 \)[/tex].
2. Set each factor to zero: [tex]\( 6y + 3 = 0 \)[/tex] or [tex]\( 4y - 1 = 0 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = -\frac{1}{2} \)[/tex] or [tex]\( y = \frac{1}{4} \)[/tex].
Since [tex]\( y = x^2 \)[/tex]:
- [tex]\( x^2 = -\frac{1}{2} \)[/tex] gives [tex]\( x = \pm \frac{\sqrt{2}}{2} i \)[/tex].
- [tex]\( x^2 = \frac{1}{4} \)[/tex] gives [tex]\( x = \pm \frac{\sqrt{3}}{6} \)[/tex].
These detailed steps explain the solution for each equation.
