The discussion focuses on solving the equation $\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$. Participants explored various approaches to find real solutions for the variables \(a\) and \(b\). The consensus highlights the importance of isolating terms and squaring both sides to eliminate the square roots, leading to a clearer path toward the solution. The collaborative effort resulted in a comprehensive understanding of the problem-solving techniques involved.
PREREQUISITESStudents, educators, and anyone interested in advanced algebraic problem-solving techniques will benefit from this discussion.
lfdahl said:My solution:
Squaring and rearranging terms yields:\[4a-b^2 = b+2+4a^2+b + 2\sqrt{(b+2)(4a^2+b)}\\\\ -(b^2+2b+1)-4(a^2-a+\frac{1}{4}) = 2\sqrt{(b+2)(4a^2+b)}\]
- under the conditions:
$4a-b^2 \geq 0 \wedge b \geq -2$ or $a \geq \left ( \frac{b}{2} \right )^2 \wedge b \geq -2$.The LHS $\le 0$:
\[-(b+1)^2-4(a-\frac{1}{2})^2=-\left ( (b+1)^2+4(a-\frac{1}{2})^2 \right ) \le 0\]The RHS $\ge 0$, so the only possibility is: $b = -1$ and $a = \frac{1}{2}$. Checking the RHS: $4a^2+b = 0$. OK.