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Question: $4x^2-3x+2=0$ has two roots $\alpha$ and $\beta$. then find the value of $(2\alpha-3)^2(2\beta-3)^2$...

4x2−3x+2=04x^2-3x+2=0 has two roots α\alpha and β\beta. then find the value of (2α−3)2(2β−3)2(2\alpha-3)^2(2\beta-3)^2

Answer

1694\frac{169}{4}

Explanation

Solution

Given 4x2−3x+2=04x^2-3x+2=0 with roots α,β\alpha, \beta. From Vieta's formulas, α+β=−(−3)/4=3/4\alpha+\beta = -(-3)/4 = 3/4 and αβ=2/4=1/2\alpha\beta = 2/4 = 1/2. The expression (2α−3)2(2β−3)2(2\alpha-3)^2(2\beta-3)^2 simplifies to [(2α−3)(2β−3)]2[(2\alpha-3)(2\beta-3)]^2. Expanding the inner product gives 4αβ−6α−6β+9=4αβ−6(α+β)+94\alpha\beta - 6\alpha - 6\beta + 9 = 4\alpha\beta - 6(\alpha+\beta) + 9. Substituting the values: 4(1/2)−6(3/4)+9=2−9/2+9=11−9/2=13/24(1/2) - 6(3/4) + 9 = 2 - 9/2 + 9 = 11 - 9/2 = 13/2. The final value is (13/2)2=169/4(13/2)^2 = 169/4.