To solve the given equation \( 2|x|^2 + 51 = |1 + 20x| \), we need to handle the absolute value expressions carefully. Let's break down the problem step by step:
- First, consider the case when \( x \geq 0 \). Then the equation becomes:
- \(2x^2 + 51 = |1 + 20x|\)
- The absolute value \(|1 + 20x|\) can take two different forms based on the sign of the expression \(1 + 20x\):
- If \(1 + 20x \geq 0 \rightarrow x \geq -\frac{1}{20}\), then we simplify to:
- \(2x^2 + 51 = 1 + 20x\)
- Rearrange the equation: \(2x^2 - 20x + 50 = 0\)
- Solve using the quadratic formula: \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -20\), and \(c = 50\).
- Calculate the discriminant: \(b^2 - 4ac = 400 - 400 = 0\)
- The solutions are: \(x = \frac{20}{4} = 5\)
- If \(1 + 20x < 0 \rightarrow x < -\frac{1}{20}\), we have a conceptual contradiction as we assumed \(x \geq 0\) initially.
- Now, consider the case when \(x < 0\). Then the equation becomes:
- \(2(-x)^2 + 51 = |1 + 20(-x)| \Rightarrow 2x^2 + 51 = |1 - 20x|\)
- Similarly, break down the absolute value \(|1 - 20x|\):
- If \(1 - 20x \geq 0 \rightarrow x \leq \frac{1}{20}\), then we simplify to:
- \(2x^2 + 51 = 1 - 20x\)
- Rearrange: \(2x^2 + 20x + 50 = 0\)
- Solve using the quadratic formula where \(a = 2\), \(b = 20\), \(c = 50\)
- Discriminant: \(b^2 - 4ac = 400 - 400 = 0\)
- The solutions are: \(x = -\frac{20}{4} = -5\)
- If \(1 - 20x < 0 \rightarrow x > \frac{1}{20}\), it doesn't need consideration here as \(x < 0\).
The valid solutions are \(x = 5\) from the first case and \(x = -5\) from the second case. Therefore, the sum of the real solutions is:
Thus, the sum of the real solutions of the given equation is 0. However, the provided options do not include this sum, indicating that the correct answer is indeed "None of these".