Answer :
Let’s tackle parts (a) and (b) step-by-step:
### Part (a): Solve [tex]\( f(x) = 0 \)[/tex] for [tex]\( f(x) = x^2(2x + 1) - 15x \)[/tex]
1. Define the function: [tex]\( f(x) = x^2(2x + 1) - 15x \)[/tex]
2. Set the function to zero:
[tex]\[ x^2(2x + 1) - 15x = 0 \][/tex]
3. Factor out the common term:
[tex]\[ x \left( x(2x + 1) - 15 \right) = 0 \][/tex]
This gives us two cases to solve: [tex]\( x = 0 \)[/tex] and [tex]\( x(2x + 1) - 15 = 0 \)[/tex].
4. Solve the first case:
[tex]\[ x = 0 \][/tex]
5. Solve the second case:
[tex]\[ x(2x + 1) - 15 = 0 \][/tex]
[tex]\[ 2x^2 + x - 15 = 0 \][/tex]
6. Use the quadratic formula: For [tex]\( ax^2 + bx + c = 0 \)[/tex], the solutions are:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -15 \)[/tex].
7. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-15) = 1 + 120 = 121 \][/tex]
8. Find the roots:
[tex]\[ x = \frac{-1 \pm \sqrt{121}}{2 \cdot 2} = \frac{-1 \pm 11}{4} \][/tex]
Therefore, the two roots are:
[tex]\[ x = \frac{10}{4} = \frac{5}{2} \quad \text{and} \quad x = \frac{-12}{4} = -3 \][/tex]
So, the solutions to [tex]\( f(x) = 0 \)[/tex] are:
[tex]\[ x = -3, \quad x = 0, \quad \text{and} \quad x = \frac{5}{2} \][/tex]
### Part (b): Solve [tex]\( y^{\frac{4}{3}}(2y^{\frac{2}{3}} + 1) - 15y^{\frac{2}{3}} = 0 \)[/tex] for [tex]\( y > 0 \)[/tex]
1. Let [tex]\( t = y^{\frac{2}{3}} \)[/tex]. Therefore, [tex]\( y = t^{\frac{3}{2}} \)[/tex].
2. Rewrite the equation in terms of [tex]\( t \)[/tex]:
[tex]\[ t^2 (2t + 1) - 15t = 0 \][/tex]
3. Factor out the common term [tex]\( t \)[/tex]:
[tex]\[ t \left( t(2t + 1) - 15 \right) = 0 \][/tex]
This gives us two cases to solve: [tex]\( t = 0 \)[/tex] and [tex]\( t(2t + 1) - 15 = 0 \)[/tex].
4. Solve the first case:
[tex]\[ t = 0 \][/tex]
However, given [tex]\( y > 0 \)[/tex], we discard [tex]\( t = 0 \)[/tex] since [tex]\( t = y^{\frac{2}{3}} \)[/tex].
5. Solve the second case:
[tex]\[ t(2t + 1) - 15 = 0 \][/tex]
[tex]\[ 2t^2 + t - 15 = 0 \][/tex]
6. Use the quadratic formula again:
[tex]\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-15)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 120}}{4} = \frac{-1 \pm \sqrt{121}}{4} \][/tex]
Therefore, the two roots are:
[tex]\[ t = \frac{10}{4} = \frac{5}{2} \quad \text{and} \quad t = \frac{-12}{4} = -3 \][/tex]
7. Consider only the positive solution:
[tex]\[ t = \frac{5}{2} \][/tex]
8. Find the corresponding [tex]\( y \)[/tex]:
[tex]\[ y = t^{\frac{3}{2}} = \left( \frac{5}{2} \right)^{\frac{3}{2}} \][/tex]
So, the solution to [tex]\( y^{\frac{4}{3}}(2y^{\frac{2}{3}} + 1) - 15y^{\frac{2}{3}} = 0 \)[/tex] for [tex]\( y > 0 \)[/tex] is:
[tex]\[ y = \left( \frac{5}{2} \right)^{\frac{3}{2}} \][/tex]
When simplified to numerical form, if desired, it is:
[tex]\[ y \approx 3.95284707521047 \][/tex]
### Part (a): Solve [tex]\( f(x) = 0 \)[/tex] for [tex]\( f(x) = x^2(2x + 1) - 15x \)[/tex]
1. Define the function: [tex]\( f(x) = x^2(2x + 1) - 15x \)[/tex]
2. Set the function to zero:
[tex]\[ x^2(2x + 1) - 15x = 0 \][/tex]
3. Factor out the common term:
[tex]\[ x \left( x(2x + 1) - 15 \right) = 0 \][/tex]
This gives us two cases to solve: [tex]\( x = 0 \)[/tex] and [tex]\( x(2x + 1) - 15 = 0 \)[/tex].
4. Solve the first case:
[tex]\[ x = 0 \][/tex]
5. Solve the second case:
[tex]\[ x(2x + 1) - 15 = 0 \][/tex]
[tex]\[ 2x^2 + x - 15 = 0 \][/tex]
6. Use the quadratic formula: For [tex]\( ax^2 + bx + c = 0 \)[/tex], the solutions are:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 2 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -15 \)[/tex].
7. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-15) = 1 + 120 = 121 \][/tex]
8. Find the roots:
[tex]\[ x = \frac{-1 \pm \sqrt{121}}{2 \cdot 2} = \frac{-1 \pm 11}{4} \][/tex]
Therefore, the two roots are:
[tex]\[ x = \frac{10}{4} = \frac{5}{2} \quad \text{and} \quad x = \frac{-12}{4} = -3 \][/tex]
So, the solutions to [tex]\( f(x) = 0 \)[/tex] are:
[tex]\[ x = -3, \quad x = 0, \quad \text{and} \quad x = \frac{5}{2} \][/tex]
### Part (b): Solve [tex]\( y^{\frac{4}{3}}(2y^{\frac{2}{3}} + 1) - 15y^{\frac{2}{3}} = 0 \)[/tex] for [tex]\( y > 0 \)[/tex]
1. Let [tex]\( t = y^{\frac{2}{3}} \)[/tex]. Therefore, [tex]\( y = t^{\frac{3}{2}} \)[/tex].
2. Rewrite the equation in terms of [tex]\( t \)[/tex]:
[tex]\[ t^2 (2t + 1) - 15t = 0 \][/tex]
3. Factor out the common term [tex]\( t \)[/tex]:
[tex]\[ t \left( t(2t + 1) - 15 \right) = 0 \][/tex]
This gives us two cases to solve: [tex]\( t = 0 \)[/tex] and [tex]\( t(2t + 1) - 15 = 0 \)[/tex].
4. Solve the first case:
[tex]\[ t = 0 \][/tex]
However, given [tex]\( y > 0 \)[/tex], we discard [tex]\( t = 0 \)[/tex] since [tex]\( t = y^{\frac{2}{3}} \)[/tex].
5. Solve the second case:
[tex]\[ t(2t + 1) - 15 = 0 \][/tex]
[tex]\[ 2t^2 + t - 15 = 0 \][/tex]
6. Use the quadratic formula again:
[tex]\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-15)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 120}}{4} = \frac{-1 \pm \sqrt{121}}{4} \][/tex]
Therefore, the two roots are:
[tex]\[ t = \frac{10}{4} = \frac{5}{2} \quad \text{and} \quad t = \frac{-12}{4} = -3 \][/tex]
7. Consider only the positive solution:
[tex]\[ t = \frac{5}{2} \][/tex]
8. Find the corresponding [tex]\( y \)[/tex]:
[tex]\[ y = t^{\frac{3}{2}} = \left( \frac{5}{2} \right)^{\frac{3}{2}} \][/tex]
So, the solution to [tex]\( y^{\frac{4}{3}}(2y^{\frac{2}{3}} + 1) - 15y^{\frac{2}{3}} = 0 \)[/tex] for [tex]\( y > 0 \)[/tex] is:
[tex]\[ y = \left( \frac{5}{2} \right)^{\frac{3}{2}} \][/tex]
When simplified to numerical form, if desired, it is:
[tex]\[ y \approx 3.95284707521047 \][/tex]
