Answer :
To solve the equation [tex]\(x^4 - 45 = -4x^2\)[/tex] and express the solutions in simplest radical form, let's go through the steps:
1. Rearrange the Equation:
Start by bringing all the terms to one side of the equation to set it equal to zero:
[tex]\[
x^4 + 4x^2 - 45 = 0
\][/tex]
2. Substitution:
Let [tex]\(y = x^2\)[/tex]. This will transform the quartic equation into a quadratic equation:
[tex]\[
y^2 + 4y - 45 = 0
\][/tex]
3. Solve the Quadratic Equation:
Use the quadratic formula to solve for [tex]\(y\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = -45\)[/tex]:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
[tex]\[
y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1}
\][/tex]
[tex]\[
y = \frac{-4 \pm \sqrt{16 + 180}}{2}
\][/tex]
[tex]\[
y = \frac{-4 \pm \sqrt{196}}{2}
\][/tex]
[tex]\[
y = \frac{-4 \pm 14}{2}
\][/tex]
This gives us the two solutions for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{10}{2} = 5 \quad \text{and} \quad y = \frac{-18}{2} = -9
\][/tex]
4. Back Substitute to Solve for [tex]\(x\)[/tex]:
- For [tex]\(y = 5\)[/tex]:
[tex]\[
x^2 = 5
\][/tex]
Thus, [tex]\(x = \pm \sqrt{5}\)[/tex].
- For [tex]\(y = -9\)[/tex]:
[tex]\[
x^2 = -9
\][/tex]
Thus, [tex]\(x = \pm \sqrt{-9} = \pm 3i\)[/tex], where [tex]\(i\)[/tex] is the imaginary unit.
5. Final Solutions:
The solutions to the equation [tex]\(x^4 - 45 = -4x^2\)[/tex] are:
[tex]\[
x = -\sqrt{5},\ \sqrt{5},\ -3i,\ 3i
\][/tex]
These are the roots expressed in simplest radical form, capturing both real and complex solutions.
1. Rearrange the Equation:
Start by bringing all the terms to one side of the equation to set it equal to zero:
[tex]\[
x^4 + 4x^2 - 45 = 0
\][/tex]
2. Substitution:
Let [tex]\(y = x^2\)[/tex]. This will transform the quartic equation into a quadratic equation:
[tex]\[
y^2 + 4y - 45 = 0
\][/tex]
3. Solve the Quadratic Equation:
Use the quadratic formula to solve for [tex]\(y\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = -45\)[/tex]:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
[tex]\[
y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-45)}}{2 \cdot 1}
\][/tex]
[tex]\[
y = \frac{-4 \pm \sqrt{16 + 180}}{2}
\][/tex]
[tex]\[
y = \frac{-4 \pm \sqrt{196}}{2}
\][/tex]
[tex]\[
y = \frac{-4 \pm 14}{2}
\][/tex]
This gives us the two solutions for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{10}{2} = 5 \quad \text{and} \quad y = \frac{-18}{2} = -9
\][/tex]
4. Back Substitute to Solve for [tex]\(x\)[/tex]:
- For [tex]\(y = 5\)[/tex]:
[tex]\[
x^2 = 5
\][/tex]
Thus, [tex]\(x = \pm \sqrt{5}\)[/tex].
- For [tex]\(y = -9\)[/tex]:
[tex]\[
x^2 = -9
\][/tex]
Thus, [tex]\(x = \pm \sqrt{-9} = \pm 3i\)[/tex], where [tex]\(i\)[/tex] is the imaginary unit.
5. Final Solutions:
The solutions to the equation [tex]\(x^4 - 45 = -4x^2\)[/tex] are:
[tex]\[
x = -\sqrt{5},\ \sqrt{5},\ -3i,\ 3i
\][/tex]
These are the roots expressed in simplest radical form, capturing both real and complex solutions.
