(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+9a2.
Step 2: Expand (a+1)2:
(a+1)2=a2+2a+1.
Step 3: Substitute the expressions into the equation:
25−30a+9a2+a=a2+2a+1−31a.
Step 4: Simplify both sides:
Left-hand side: 9a2−29a+25.
Right-hand side: a2−29a+1.
Set the equation 9a2−29a+25=a2−29a+1.
Simplify the equation:
Subtract a2−29a+1 from both sides:
9a2−a2−29a+29a+25−1=0.
8a2+24=0.
8a2=−24.
Divide through by 8:
a2=−3.
Since a2=−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.