Answer :
To solve the equation [tex]\(x^4 - 45x^2 + 324 = 0\)[/tex] using the substitution method, follow these steps:
1. Substitution: Let [tex]\( y = x^2 \)[/tex]. This transforms the equation into a quadratic equation: [tex]\( y^2 - 45y + 324 = 0 \)[/tex].
2. Solve the Quadratic Equation: We need to find the values of [tex]\( y \)[/tex] that satisfy [tex]\( y^2 - 45y + 324 = 0 \)[/tex]. This can be done using the quadratic formula:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -45 \)[/tex], and [tex]\( c = 324 \)[/tex].
3. Discriminant: Calculate the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = (-45)^2 - 4 \times 1 \times 324 = 2025 - 1296 = 729
\][/tex]
Since the discriminant is positive, there are two real solutions for [tex]\( y \)[/tex].
4. Find [tex]\( y \)[/tex] Values:
- The first solution for [tex]\( y \)[/tex] is:
[tex]\[
y_1 = \frac{-(-45) + \sqrt{729}}{2 \times 1} = \frac{45 + 27}{2} = \frac{72}{2} = 36
\][/tex]
- The second solution for [tex]\( y \)[/tex] is:
[tex]\[
y_2 = \frac{-(-45) - \sqrt{729}}{2 \times 1} = \frac{45 - 27}{2} = \frac{18}{2} = 9
\][/tex]
5. Back Substitute to Find [tex]\( x \)[/tex]:
- For [tex]\( y = 36 \)[/tex]: since [tex]\( y = x^2 \)[/tex], [tex]\( x^2 = 36 \)[/tex]. So, [tex]\( x = \sqrt{36} = 6 \)[/tex] or [tex]\( x = -\sqrt{36} = -6 \)[/tex].
- For [tex]\( y = 9 \)[/tex]: since [tex]\( y = x^2 \)[/tex], [tex]\( x^2 = 9 \)[/tex]. So, [tex]\( x = \sqrt{9} = 3 \)[/tex] or [tex]\( x = -\sqrt{9} = -3 \)[/tex].
6. Check for Extraneous Solutions: All the roots [tex]\( x = 6, -6, 3, -3 \)[/tex] need to be checked against the original equation [tex]\( x^4 - 45x^2 + 324 = 0 \)[/tex]. Each of these values satisfies the equation.
Thus, the solutions to the equation [tex]\(x^4 - 45x^2 + 324 = 0\)[/tex] are [tex]\( x = 6, -6, 3, -3 \)[/tex], and there are no extraneous solutions.
1. Substitution: Let [tex]\( y = x^2 \)[/tex]. This transforms the equation into a quadratic equation: [tex]\( y^2 - 45y + 324 = 0 \)[/tex].
2. Solve the Quadratic Equation: We need to find the values of [tex]\( y \)[/tex] that satisfy [tex]\( y^2 - 45y + 324 = 0 \)[/tex]. This can be done using the quadratic formula:
[tex]\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -45 \)[/tex], and [tex]\( c = 324 \)[/tex].
3. Discriminant: Calculate the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[
\Delta = b^2 - 4ac = (-45)^2 - 4 \times 1 \times 324 = 2025 - 1296 = 729
\][/tex]
Since the discriminant is positive, there are two real solutions for [tex]\( y \)[/tex].
4. Find [tex]\( y \)[/tex] Values:
- The first solution for [tex]\( y \)[/tex] is:
[tex]\[
y_1 = \frac{-(-45) + \sqrt{729}}{2 \times 1} = \frac{45 + 27}{2} = \frac{72}{2} = 36
\][/tex]
- The second solution for [tex]\( y \)[/tex] is:
[tex]\[
y_2 = \frac{-(-45) - \sqrt{729}}{2 \times 1} = \frac{45 - 27}{2} = \frac{18}{2} = 9
\][/tex]
5. Back Substitute to Find [tex]\( x \)[/tex]:
- For [tex]\( y = 36 \)[/tex]: since [tex]\( y = x^2 \)[/tex], [tex]\( x^2 = 36 \)[/tex]. So, [tex]\( x = \sqrt{36} = 6 \)[/tex] or [tex]\( x = -\sqrt{36} = -6 \)[/tex].
- For [tex]\( y = 9 \)[/tex]: since [tex]\( y = x^2 \)[/tex], [tex]\( x^2 = 9 \)[/tex]. So, [tex]\( x = \sqrt{9} = 3 \)[/tex] or [tex]\( x = -\sqrt{9} = -3 \)[/tex].
6. Check for Extraneous Solutions: All the roots [tex]\( x = 6, -6, 3, -3 \)[/tex] need to be checked against the original equation [tex]\( x^4 - 45x^2 + 324 = 0 \)[/tex]. Each of these values satisfies the equation.
Thus, the solutions to the equation [tex]\(x^4 - 45x^2 + 324 = 0\)[/tex] are [tex]\( x = 6, -6, 3, -3 \)[/tex], and there are no extraneous solutions.
