\(6(2 - \sqrt{2})\)
We are given the equation: \[ x^2 + |2x - 3| - 4 = 0 \]
Case 1: \( x \geq \frac{3}{2} \)
The absolute value simplifies to: \[ x^2 + 2x - 3 - 4 = 0 \] Simplified equation: \[ x^2 + 2x - 7 = 0 \] Solution: \[ x = 2\sqrt{2} - 1 \]
Case 2: \( x < \frac{3}{2} \)
The absolute value becomes: \[ x^2 + 3 - 2x - 4 = 0 \] Simplified equation: \[ x^2 - 2x - 1 = 0 \] Solution: \[ x = 1 - \sqrt{2} \]
Sum of Squares Calculation:
\[ \left( 2\sqrt{2} - 1 \right)^2 + \left( 1 - \sqrt{2} \right)^2 \] Expansion: \[ = 8 - 4\sqrt{2} + 1 + 1 - 2\sqrt{2} + 2 \] Simplification: \[ = 12 - 6\sqrt{2} = 6(2 - \sqrt{2}) \]
Final Answer:
The sum of squares of the solutions is \(6(2 - \sqrt{2})\).
Let p and q be two real numbers such that p + q = 3 and p4 + q4 = 369. Then
\((\frac{1}{p} + \frac{1}{q} )^{-2}\)
is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)