Answer :
To solve the equation [tex]\( 5y = 40y^{\frac{1}{2}} \)[/tex], let's go through the solution step by step.
1. Rewrite the equation:
[tex]\[ 5y = 40y^{\frac{1}{2}} \][/tex]
2. Isolate the square root term:
Divide both sides of the equation by 5 to simplify:
[tex]\[ y = 8y^{\frac{1}{2}} \][/tex]
3. Introduce a substitution:
Let [tex]\( z = y^{\frac{1}{2}} \)[/tex]. Therefore, [tex]\( y = z^2 \)[/tex].
4. Substitute [tex]\( z \)[/tex] back into the equation:
Substituting [tex]\( y = z^2 \)[/tex]:
[tex]\[ z^2 = 8z \][/tex]
5. Solve for [tex]\( z \)[/tex]:
Rearrange this into a standard quadratic equation form:
[tex]\[ z^2 - 8z = 0 \][/tex]
Factor out a common term:
[tex]\[ z(z - 8) = 0 \][/tex]
This gives us two potential solutions:
[tex]\[ z = 0 \quad \text{or} \quad z - 8 = 0 \implies z = 8 \][/tex]
6. Re-substitute [tex]\( z \)[/tex] with [tex]\( y^{\frac{1}{2}} \)[/tex]:
Remembering that [tex]\( z = y^{\frac{1}{2}} \)[/tex]:
[tex]\[ y^{\frac{1}{2}} = 0 \quad \text{or} \quad y^{\frac{1}{2}} = 8 \][/tex]
7. Square both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y^{\frac{1}{2}} = 0 \implies y = 0^2 = 0 \][/tex]
[tex]\[ y^{\frac{1}{2}} = 8 \implies y = 8^2 = 64 \][/tex]
8. Conclusion:
Therefore, the solutions to the equation [tex]\( 5y = 40y^{\frac{1}{2}} \)[/tex] are
[tex]\[ y = 0 \quad \text{and} \quad y = 64 \][/tex]
So, the final answers are [tex]\( y = 0 \)[/tex] and [tex]\( y = 64 \)[/tex].
1. Rewrite the equation:
[tex]\[ 5y = 40y^{\frac{1}{2}} \][/tex]
2. Isolate the square root term:
Divide both sides of the equation by 5 to simplify:
[tex]\[ y = 8y^{\frac{1}{2}} \][/tex]
3. Introduce a substitution:
Let [tex]\( z = y^{\frac{1}{2}} \)[/tex]. Therefore, [tex]\( y = z^2 \)[/tex].
4. Substitute [tex]\( z \)[/tex] back into the equation:
Substituting [tex]\( y = z^2 \)[/tex]:
[tex]\[ z^2 = 8z \][/tex]
5. Solve for [tex]\( z \)[/tex]:
Rearrange this into a standard quadratic equation form:
[tex]\[ z^2 - 8z = 0 \][/tex]
Factor out a common term:
[tex]\[ z(z - 8) = 0 \][/tex]
This gives us two potential solutions:
[tex]\[ z = 0 \quad \text{or} \quad z - 8 = 0 \implies z = 8 \][/tex]
6. Re-substitute [tex]\( z \)[/tex] with [tex]\( y^{\frac{1}{2}} \)[/tex]:
Remembering that [tex]\( z = y^{\frac{1}{2}} \)[/tex]:
[tex]\[ y^{\frac{1}{2}} = 0 \quad \text{or} \quad y^{\frac{1}{2}} = 8 \][/tex]
7. Square both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y^{\frac{1}{2}} = 0 \implies y = 0^2 = 0 \][/tex]
[tex]\[ y^{\frac{1}{2}} = 8 \implies y = 8^2 = 64 \][/tex]
8. Conclusion:
Therefore, the solutions to the equation [tex]\( 5y = 40y^{\frac{1}{2}} \)[/tex] are
[tex]\[ y = 0 \quad \text{and} \quad y = 64 \][/tex]
So, the final answers are [tex]\( y = 0 \)[/tex] and [tex]\( y = 64 \)[/tex].