Answer :

Sure! Let's solve the given biquadratic equations step-by-step.

### a) Solve [tex]\(x^4 - 3x^2 - 4 = 0\)[/tex]

1. Substitution: Let [tex]\(y = x^2\)[/tex]. Then [tex]\(x^4 = y^2\)[/tex]. Substitute into the equation:
[tex]\[
y^2 - 3y - 4 = 0
\][/tex]

2. Solve the quadratic: The equation is now a quadratic equation in terms of [tex]\(y\)[/tex]. We can use the quadratic formula:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = -4\)[/tex].

3. Quadratic formula results:
[tex]\[
y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}
\][/tex]
[tex]\[
y = \frac{3 \pm \sqrt{9 + 16}}{2}
\][/tex]
[tex]\[
y = \frac{3 \pm \sqrt{25}}{2}
\][/tex]
[tex]\[
y = \frac{3 \pm 5}{2}
\][/tex]

4. Solving for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{8}{2} = 4 \quad \text{and} \quad y = \frac{-2}{2} = -1
\][/tex]

5. Substitute back for [tex]\(x\)[/tex]:
- For [tex]\(y = 4\)[/tex], [tex]\(x^2 = 4\)[/tex] gives [tex]\(x = \pm 2\)[/tex].
- For [tex]\(y = -1\)[/tex], [tex]\(x^2 = -1\)[/tex] has no real solutions (results in complex numbers).

Therefore, the solutions for [tex]\(x\)[/tex] are [tex]\(x = 2\)[/tex] and [tex]\(x = -2\)[/tex].

### b) Solve [tex]\(4x^4 + 21x^2 - 25 = 0\)[/tex]

1. Substitution: Let [tex]\(y = x^2\)[/tex]. Then [tex]\(x^4 = y^2\)[/tex]. Substitute into the equation:
[tex]\[
4y^2 + 21y - 25 = 0
\][/tex]

2. Using the quadratic formula for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\(a = 4\)[/tex], [tex]\(b = 21\)[/tex], and [tex]\(c = -25\)[/tex].

3. Quadratic formula results:
[tex]\[
y = \frac{-21 \pm \sqrt{21^2 - 4 \cdot 4 \cdot (-25)}}{2 \cdot 4}
\][/tex]
[tex]\[
y = \frac{-21 \pm \sqrt{441 + 400}}{8}
\][/tex]
[tex]\[
y = \frac{-21 \pm \sqrt{841}}{8}
\][/tex]
[tex]\[
y = \frac{-21 \pm 29}{8}
\][/tex]

4. Solving for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{8}{8} = 1 \quad \text{and} \quad y = \frac{-50}{8} = -6.25
\][/tex]

5. Substitute back for [tex]\(x\)[/tex]:
- For [tex]\(y = 1\)[/tex], [tex]\(x^2 = 1\)[/tex] gives [tex]\(x = \pm 1\)[/tex].
- For [tex]\(y = -6.25\)[/tex], [tex]\(x^2 = -6.25\)[/tex] has no real solutions (results in complex numbers).

Therefore, the solutions for [tex]\(x\)[/tex] are [tex]\(x = 1\)[/tex] and [tex]\(x = -1\)[/tex].

To summarize:
- The solutions for equation (a) are [tex]\(x = 2\)[/tex] and [tex]\(x = -2\)[/tex].
- The solutions for equation (b) are [tex]\(x = 1\)[/tex] and [tex]\(x = -1\)[/tex].

Thanks for taking the time to read 1 Resuelve las siguientes ecuaciones bicuadráticas a tex x 4 3x 2 4 0 tex b tex 4x 4 21x 2 25 0 tex. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada