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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