Mathematics 2004 JAMB Past Questions - Question 31: find the value of α² β² if α + β = 2 and the distance between the points (1, α and (β, 1) is 3 units

Correct Answer

C

Explanation

To find the value of α²β², we need to use the given information and solve for α and β.We are given that α + β = 2 and the distance between the points (1, α) and (β, 1) is 3 units.Using the distance formula, we can calculate the distance between the two points:√((β - 1)² + (α - 1)²) = 3Simplifying the equation:(β - 1)² + (α - 1)² = 9Expanding and rearranging the equation:β² - 2β + 1 + α² - 2α + 1 = 9β² + α² - 2β - 2α - 7 = 0Now, we can use the given equation α + β = 2 to solve for α or β. Let's solve for α:α = 2 - βSubstituting this value into the equation β² + α² - 2β - 2α - 7 = 0:β² + (2 - β)² - 2β - 2(2 - β) - 7 = 0Expanding and simplifying:β² + 4 - 4β + β² - 2β - 4 + 2β - 7 = 02β² - 8β - 7 = 0Now, we can solve this quadratic equation for β using the quadratic formula:β = (-(-8) ± √((-8)² - 4(2)(-7))) / (2(2))Simplifying further:β = (8 ± √(64 + 56)) / 4β = (8 ± √120) / 4β = (8 ± 2√30) / 4β = 2 ± 0.5√30Now, we can substitute these values of β back into the equation α + β = 2 to solve for α:α = 2 - βFor β = 2 + 0.5√30:α = 2 - (2 + 0.5√30)α = -0.5√30For β = 2 - 0.5√30:α = 2 - (2 - 0.5√30)α = 0.5√30Now, we can calculate the value of α²β²:α²β² = (-0.5√30)² * (0.5√30)²α²β² = 0.25 * 30α²β² = 7.5Therefore, the value of α²β² is 7.5.