Please gelp me wirh algerba 2 work

Question

The volume of a figure is x³--9x. The surface area of another figure is 8x².Disregarding the units,the volume of the first figure equals the surface area of the second figure. What are the possible values of x

Answer 1

Unless I'm missing something, this seems like an unnecessarily complicated way to get you to solve the equation: x³-9x = 8x² 😅

When you are solving equations like this, the first step is to set the equation equal to zero by moving all of the terms to one side.

Step 0 (Given Equation): x³ - 9x = 8x²

Step 1 (Move all terms to one side): x³ - 8x² - 9x = 0 (Notice that I reordered the terms so that the powers of x are decreasing.)

Step 2 (Factor out the "x"): x(x² - 8x - 9) = 0

Step 3 (Factor the trinomial inside the parentheses): x(x - 9)(x + 1)=0

* Now, to solve for x, set each factor equal to 0. That gives us 3 equations:

Equations: x = 0, x - 9 = 0 x + 1 = 0

Solutions: x = 0, x = 9, x = -1

Now, this is where it's helpful to substitute the values back in the original expressions to both check your solutions and to see if they make sense:

VOLUME = x³-9x

x = 0. --> (0)³-9(0) = 0

x = 9 --> (9)³-9(9) = 648

x = -1 --> (-1)³-9(-1) = 8

SURFACE AREA = 8x²

x = 0. --> 8(0)² = 0

x = 9 --> 8(9)² = 648

x = -1 --> 8(-1)² = 8

So, on the bright side, all of the volumes are equal to the surface area values, which is what should happen! The one tricky answer though, is x = 0. If the volume and surface area is 0, technically you wouldn't have a shape, so... the answers that are acceptable would be x = 9 and x = -1.

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