Answer :
To solve the equation [tex]\(x^4 - 14x^2 + 45 = 0\)[/tex], we'll use a substitution method to simplify it into a quadratic equation. Here's a clear step-by-step solution:
1. Substitute [tex]\(y = x^2\)[/tex]:
By letting [tex]\(y = x^2\)[/tex], the original equation [tex]\(x^4 - 14x^2 + 45 = 0\)[/tex] becomes:
[tex]\[ y^2 - 14y + 45 = 0 \][/tex]
2. Solve the quadratic equation for [tex]\(y\)[/tex]:
The equation [tex]\(y^2 - 14y + 45 = 0\)[/tex] is a standard quadratic equation of the form [tex]\(ay^2 + by + c = 0\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -14\)[/tex], and [tex]\(c = 45\)[/tex].
To find the solutions for [tex]\(y\)[/tex], use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Calculate the discriminant:
[tex]\[ b^2 - 4ac = (-14)^2 - 4 \times 1 \times 45 = 196 - 180 = 16 \][/tex]
Substitute the values into the quadratic formula:
[tex]\[ y_1 = \frac{-(-14) + \sqrt{16}}{2 \times 1} = \frac{14 + 4}{2} = \frac{18}{2} = 9 \][/tex]
[tex]\[ y_2 = \frac{-(-14) - \sqrt{16}}{2 \times 1} = \frac{14 - 4}{2} = \frac{10}{2} = 5 \][/tex]
3. Solve for [tex]\(x\)[/tex] using [tex]\(y = x^2\)[/tex]:
Since [tex]\(y = x^2\)[/tex], we find [tex]\(x\)[/tex] by taking the square roots of each solution for [tex]\(y\)[/tex].
- For [tex]\(y = 9\)[/tex]:
[tex]\[ x = \sqrt{9} = 3 \quad \text{and} \quad x = -\sqrt{9} = -3 \][/tex]
- For [tex]\(y = 5\)[/tex]:
[tex]\[ x = \sqrt{5} \approx 2.236 \quad \text{and} \quad x = -\sqrt{5} \approx -2.236 \][/tex]
4. Final solutions for [tex]\(x\)[/tex]:
The four solutions for the equation [tex]\(x^4 - 14x^2 + 45 = 0\)[/tex] are:
[tex]\[ x = 3, \; x = -3, \; x = \sqrt{5}, \; x = -\sqrt{5} \][/tex]
Approximately, these are:
[tex]\[ x = 3.0, \; x = -3.0, \; x \approx 2.236, \; x \approx -2.236 \][/tex]
These steps show how to solve the equation by simplifying it to a quadratic form, finding the roots of the quadratic, and then determining the corresponding values of [tex]\(x\)[/tex].
1. Substitute [tex]\(y = x^2\)[/tex]:
By letting [tex]\(y = x^2\)[/tex], the original equation [tex]\(x^4 - 14x^2 + 45 = 0\)[/tex] becomes:
[tex]\[ y^2 - 14y + 45 = 0 \][/tex]
2. Solve the quadratic equation for [tex]\(y\)[/tex]:
The equation [tex]\(y^2 - 14y + 45 = 0\)[/tex] is a standard quadratic equation of the form [tex]\(ay^2 + by + c = 0\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -14\)[/tex], and [tex]\(c = 45\)[/tex].
To find the solutions for [tex]\(y\)[/tex], use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Calculate the discriminant:
[tex]\[ b^2 - 4ac = (-14)^2 - 4 \times 1 \times 45 = 196 - 180 = 16 \][/tex]
Substitute the values into the quadratic formula:
[tex]\[ y_1 = \frac{-(-14) + \sqrt{16}}{2 \times 1} = \frac{14 + 4}{2} = \frac{18}{2} = 9 \][/tex]
[tex]\[ y_2 = \frac{-(-14) - \sqrt{16}}{2 \times 1} = \frac{14 - 4}{2} = \frac{10}{2} = 5 \][/tex]
3. Solve for [tex]\(x\)[/tex] using [tex]\(y = x^2\)[/tex]:
Since [tex]\(y = x^2\)[/tex], we find [tex]\(x\)[/tex] by taking the square roots of each solution for [tex]\(y\)[/tex].
- For [tex]\(y = 9\)[/tex]:
[tex]\[ x = \sqrt{9} = 3 \quad \text{and} \quad x = -\sqrt{9} = -3 \][/tex]
- For [tex]\(y = 5\)[/tex]:
[tex]\[ x = \sqrt{5} \approx 2.236 \quad \text{and} \quad x = -\sqrt{5} \approx -2.236 \][/tex]
4. Final solutions for [tex]\(x\)[/tex]:
The four solutions for the equation [tex]\(x^4 - 14x^2 + 45 = 0\)[/tex] are:
[tex]\[ x = 3, \; x = -3, \; x = \sqrt{5}, \; x = -\sqrt{5} \][/tex]
Approximately, these are:
[tex]\[ x = 3.0, \; x = -3.0, \; x \approx 2.236, \; x \approx -2.236 \][/tex]
These steps show how to solve the equation by simplifying it to a quadratic form, finding the roots of the quadratic, and then determining the corresponding values of [tex]\(x\)[/tex].
