Answer:
Step-by-step explanation:
An equation is solved for one of the variables by undoing the operations done on that variable. Here, we can solve for 'a' and solve for 'b' to find the desired values.
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value of a
Solve for 'a', then substitute the value of 'b'.
[tex]\sqrt{9a}=\dfrac{b^2-3}{2}\\\\9a=\left(\dfrac{b^2-3}{2}\right)^2=\dfrac{(b^2-3)^2}{4}\\\\a=\dfrac{(b^2-3)^2}{36}=\dfrac{((-3)^2-3)^2}{36}=\dfrac{6^2}{36}\\\\\boxed{a=1}[/tex]
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value of b
Solve for 'b', then substitute the value of 'a'.
[tex]\sqrt{9a}=\dfrac{b^2-3}{2}\\\\2\cdot3\sqrt{a}=b^2-3\\\\6\sqrt{a}+3=b^2\\\\b=\sqrt{6\sqrt{a}+3}=\sqrt{6\sqrt{4}+3}=\sqrt{12+3}\\\\\boxed{b=\sqrt{15}}[/tex]
