Given: \(\frac{x^2+4}{5}=\frac{x+8}{3}\)

Cross multiply: \((x^2+4)(3) = (5)(x+8)\)

Expand: \(3x^2 + 12 = 5x + 40\)

Subtract \(5x\) from both sides of the equation: \(3x^2 - 5x + 12 = 40\)

Subtract \(40\) from both sides of the equation: \(3x^2 - 5x - 28 = 0\)

Factor: \((3x + 7)(x - 4) = 0\)

So, EITHER \((3x + 7) = 0\) OR \((x - 4) = 0\)

If \((3x + 7) = 0\), then \(x = \frac{-7}{3}\), in which case \(x^2\) is some POSITIVE number
In this case, Quantity B is greater.

If \((x -4) = 0\), then \(x = 4\), in which case \(x^2 = 16\)
In this case, Quantity B is greater.

In BOTH possible cases, Quantity B is greater, so the correct answer is B

Cheers,
Brent

For this kind of questions after finding the answers substitute the values in equation and check

by making RHS to 0 we get a equation like this
\(3x^2-5x-28=0\)

by solving the equation we will get x=4 and x=(-7/3)

susbstituting these values in above equation
when x=4
\((4^2+4)/5 = (4+8)/3\)
4=4 so x=4 is one solution
quan A=x=4
quan B=x^2=16
quan A < quan B

when x=(-7/3)
(17/9)= (17/9) x=(-7/3) is also solution
quan A=x= (-7/3)
quan B=x^2=(49/9)
quan A < quan B

so option B is correct

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