Answer :
To solve the polynomial equation [tex]\(x^4 - 4x^2 + 3 = 0\)[/tex], we can start by making a substitution to simplify the equation. Let's set [tex]\(y = x^2\)[/tex]. This transforms our original equation into a quadratic equation in terms of [tex]\(y\)[/tex]:
[tex]\[ y^2 - 4y + 3 = 0 \][/tex]
Next, we solve this quadratic equation for [tex]\(y\)[/tex]. The standard form of a quadratic equation is [tex]\(ay^2 + by + c = 0\)[/tex]. In our case, [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 3\)[/tex]. We can find the solutions using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{16 - 12}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 2}{2} \][/tex]
This gives us two solutions for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{4 + 2}{2} = 3 \quad \text{and} \quad y = \frac{4 - 2}{2} = 1 \][/tex]
Now, recall our substitution [tex]\(y = x^2\)[/tex]. We have [tex]\(x^2 = 3\)[/tex] and [tex]\(x^2 = 1\)[/tex]. We solve these equations to find [tex]\(x\)[/tex]:
For [tex]\(x^2 = 3\)[/tex]:
[tex]\[ x = \pm \sqrt{3} \][/tex]
For [tex]\(x^2 = 1\)[/tex]:
[tex]\[ x = \pm 1 \][/tex]
Thus, the complete set of solutions for the original equation [tex]\(x^4 - 4x^2 + 3 = 0\)[/tex] is:
[tex]\[ x = -\sqrt{3}, -1, 1, \sqrt{3} \][/tex]
[tex]\[ y^2 - 4y + 3 = 0 \][/tex]
Next, we solve this quadratic equation for [tex]\(y\)[/tex]. The standard form of a quadratic equation is [tex]\(ay^2 + by + c = 0\)[/tex]. In our case, [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 3\)[/tex]. We can find the solutions using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(3)}}{2(1)} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{16 - 12}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 2}{2} \][/tex]
This gives us two solutions for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{4 + 2}{2} = 3 \quad \text{and} \quad y = \frac{4 - 2}{2} = 1 \][/tex]
Now, recall our substitution [tex]\(y = x^2\)[/tex]. We have [tex]\(x^2 = 3\)[/tex] and [tex]\(x^2 = 1\)[/tex]. We solve these equations to find [tex]\(x\)[/tex]:
For [tex]\(x^2 = 3\)[/tex]:
[tex]\[ x = \pm \sqrt{3} \][/tex]
For [tex]\(x^2 = 1\)[/tex]:
[tex]\[ x = \pm 1 \][/tex]
Thus, the complete set of solutions for the original equation [tex]\(x^4 - 4x^2 + 3 = 0\)[/tex] is:
[tex]\[ x = -\sqrt{3}, -1, 1, \sqrt{3} \][/tex]
