Solve the Equation: Balancing (5-3a)² + a and (a+1)² - 31a

$(5-3a)^2+a=(a+1)^2-31a$

To solve this problem, we'll follow these steps:

• Step 1: Expand $(5-3a)^2$.
• Step 2: Expand $(a+1)^2$.
• Step 3: Combine like terms and set the equation to zero.
• Step 4: Solve for $a$.

Now, let's work through each step:

Step 1: Expand $(5-3a)^2$:

$(5-3a)^2 = 25 - 30a + 9a^2$.

Step 2: Expand $(a+1)^2$:

$(a+1)^2 = a^2 + 2a + 1$.

Step 3: Substitute the expressions into the equation:

$25 - 30a + 9a^2 + a = a^2 + 2a + 1 - 31a$.

Step 4: Simplify both sides:

Left-hand side: $9a^2 - 29a + 25$.

Right-hand side: $a^2 - 29a + 1$.

Set the equation $9a^2 - 29a + 25 = a^2 - 29a + 1$.

Simplify the equation:

Subtract $a^2 - 29a + 1$ from both sides:

$9a^2 - a^2 - 29a + 29a + 25 - 1 = 0$.

$8a^2 + 24 = 0$.

$8a^2 = -24$.

Divide through by 8:

$a^2 = -3$.

Since $a^2 = -3$, there are no real solutions for $a$ because no real number squared equals a negative number. Thus, there are no solutions in the real number set.

Therefore, the correct answer is No solution.