Find the square root of (2.25)^{2} + 16 - 4(2 | vedclass

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Question

(C) Let $a = 2.25$ and $b = 4$.The expression is $a^2 + b^2 - ab - \frac{a^3}{a+b}$.Simplifying the expression:$k = \frac{(a^2 + b^2 - ab)(a+b) - a^3}

Vedclass · Accepted Answer

$3.2$

Vedclass · Answer

(C) Let $a = 2.25$ and $b = 4$.The expression is $a^2 + b^2 - ab - \frac{a^3}{a+b}$.Simplifying the expression:$k = \frac{(a^2 + b^2 - ab)(a+b) - a^3}{a+b}$Using the algebraic identity $(a+b)(a^2 + b^2 - ab) = a^3 + b^3$:$k = \frac{(a^3 + b^3) - a^3}{a+b} = \frac{b^3}{a+b}$Substituting the values $a = 2.25 = \frac{9}{4}$ and $b = 4$:$k = \frac{4^3}{\frac{9}{4} + 4} = \frac{64}{\frac{9+16}{4}} = \frac{64}{\frac{25}{4}} = \frac{64 	imes 4}{25} = \frac{256}{25}$The square root is $\sqrt{k} = \sqrt{\frac{256}{25}} = \frac{16}{5} = 3.2$.

Find the square root of $(2.25)^{2} + 16 - 4(2.25) - \frac{(2.25)^{3}}{2.25 + 4} ?$

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