1. **State the problem:** We are given a triangle with sides labeled $a^2 - 10$, $a^2 - 5a$, and $6$. We want to find the value(s) of $a$ that satisfy the triangle's conditions. 2. **Identify the problem type:** Since the triangle sides are algebraic expressions, we can use the Pythagorean theorem if the triangle is right-angled, or check for triangle inequality if not specified. 3. **Assuming the triangle is right-angled:** The longest side should be the hypotenuse. Let's check which side is longest by comparing expressions for possible $a$ values. 4. **Set up the Pythagorean theorem:** Suppose $6$ is the hypotenuse, then $$ (a^2 - 10)^2 + (a^2 - 5a)^2 = 6^2 $$ 5. **Expand and simplify:** $$ (a^2 - 10)^2 + (a^2 - 5a)^2 = 36 $$ $$ (a^2)^2 - 2 \times 10 \times a^2 + 10^2 + (a^2)^2 - 2 \times 5a \times a^2 + (5a)^2 = 36 $$ $$ a^4 - 20a^2 + 100 + a^4 - 10a^3 + 25a^2 = 36 $$ 6. **Combine like terms:** $$ 2a^4 - 10a^3 + 5a^2 + 100 = 36 $$ 7. **Bring all terms to one side:** $$ 2a^4 - 10a^3 + 5a^2 + 64 = 0 $$ 8. **Solve the quartic equation:** This is complex; alternatively, check if $a^2 - 10$ or $a^2 - 5a$ could be the hypotenuse. 9. **Try $a^2 - 10$ as hypotenuse:** $$ 6^2 + (a^2 - 5a)^2 = (a^2 - 10)^2 $$ $$ 36 + (a^2 - 5a)^2 = (a^2 - 10)^2 $$ 10. **Expand:** $$ 36 + a^4 - 10a^3 + 25a^2 = a^4 - 20a^2 + 100 $$ 11. **Simplify:** $$ 36 + a^4 - 10a^3 + 25a^2 - a^4 + 20a^2 - 100 = 0 $$ $$ -10a^3 + 45a^2 - 64 = 0 $$ 12. **Solve cubic:** $$ -10a^3 + 45a^2 - 64 = 0 $$ 13. **Try $a^2 - 5a$ as hypotenuse:** $$ 6^2 + (a^2 - 10)^2 = (a^2 - 5a)^2 $$ $$ 36 + (a^2 - 10)^2 = (a^2 - 5a)^2 $$ 14. **Expand:** $$ 36 + a^4 - 20a^2 + 100 = a^4 - 10a^3 + 25a^2 $$ 15. **Simplify:** $$ 136 - 20a^2 = -10a^3 + 25a^2 $$ $$ 0 = -10a^3 + 45a^2 - 136 $$ 16. **Summary:** The possible equations to solve are: $$ -10a^3 + 45a^2 - 64 = 0 $$ $$ -10a^3 + 45a^2 - 136 = 0 $$ 17. **Solve for $a$ numerically or graphically to find valid $a$ values.** **Final answer:** The values of $a$ satisfy either $$ -10a^3 + 45a^2 - 64 = 0 $$ or $$ -10a^3 + 45a^2 - 136 = 0 $$ These can be solved using numerical methods.