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Question 58
If $$A = \frac{3 \times 4 + 2}{6 \div 3 + 2}$$ and $$B = \frac{12 \div 6 \times 2}{8 \div 4 - 3}$$, then what is the value of $$A^2 + B^2 - 2AB$$?
Solution
$$A = \frac{3 \times 4 + 2}{6 \div 3 + 2}$$
$$A=\frac{12+2}{2+2}$$
$$A=\frac{14}{4}$$A = 3.5
$$B = \frac{12 \div 6 \times 2}{8 \div 4 - 3}$$
$$B=\frac{\frac{12}{6}\times2}{\frac{8}{4}-3}$$
$$B=\frac{2\times2}{2-3}$$B = -4
Value of $$A^2 + B^2 - 2AB$$ =Â $$\left(A-B\right)^2$$
= $$(3.5-(-4))^2$$
= $$(3.5+4)^2$$
= $$(7.5)^2$$
=Â 56.25
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