JAMB - Mathematics (2004 - No. 2)
Find the value of α\(^2\) + β\(^2\) if α + β = 2 and the distance between points (1, α) and (β, 1) is 3 units
14
3
5
11
स्पष्टीकरण
PQ = \(\sqrt{(β - 1)^{2} + (1 - α)^{2}}\)
3 = \(\sqrt{(β^{2} -2β^{2} + 1 + 1 - 2α + α^{2})}\)
3 = \(\sqrt{(α^{2} + β^{2} - 2α + 2β + 2)}\)
3 = \(\sqrt{(α^{2} + β^{2} - 2(α + β) + 2)}\)
3 = \(\sqrt{(α^{2} + β^{2} - 2 * 2 + 2)}\)
3 = \(\sqrt{(α^{2} + β^{2} - 2)}\)
9 = \(α^{2} + β^{2} - 2)\)
\(α^{2} + β^{2}\) = 9 + 2
\(α^{2} + β^{2}\) = 11
3 = \(\sqrt{(β^{2} -2β^{2} + 1 + 1 - 2α + α^{2})}\)
3 = \(\sqrt{(α^{2} + β^{2} - 2α + 2β + 2)}\)
3 = \(\sqrt{(α^{2} + β^{2} - 2(α + β) + 2)}\)
3 = \(\sqrt{(α^{2} + β^{2} - 2 * 2 + 2)}\)
3 = \(\sqrt{(α^{2} + β^{2} - 2)}\)
9 = \(α^{2} + β^{2} - 2)\)
\(α^{2} + β^{2}\) = 9 + 2
\(α^{2} + β^{2}\) = 11