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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 the quadratic equation 4x2−3x+2=04x^2 - 3x + 2 = 0, by Vieta's formulas: Sum of roots: α+β=−(−3)/4=3/4\alpha + \beta = -(-3)/4 = 3/4 Product of roots: αβ=2/4=1/2\alpha\beta = 2/4 = 1/2

The expression to evaluate is (2α−3)2(2β−3)2=[(2α−3)(2β−3)]2(2\alpha-3)^2(2\beta-3)^2 = [(2\alpha-3)(2\beta-3)]^2.

Expand the inner product: (2α−3)(2β−3)=4αβ−6α−6β+9(2\alpha-3)(2\beta-3) = 4\alpha\beta - 6\alpha - 6\beta + 9 =4αβ−6(α+β)+9= 4\alpha\beta - 6(\alpha + \beta) + 9

Substitute the values of α+β\alpha + \beta and αβ\alpha\beta: =4(1/2)−6(3/4)+9= 4(1/2) - 6(3/4) + 9 =2−9/2+9= 2 - 9/2 + 9 =11−9/2= 11 - 9/2 =(22−9)/2=13/2= (22 - 9)/2 = 13/2

Now, square this result: [(2α−3)(2β−3)]2=(13/2)2=169/4[(2\alpha-3)(2\beta-3)]^2 = (13/2)^2 = 169/4.