If each side of a regular pentagon is x^2 + 17 and each side of a regular hexagon is x^2 + 2x, for what value or values of x will the perimeters of the figures be equal?

When is

6(x^2+2x) = 5(x^2+17) ?

solve as a quadratic

Deebbii naafe rgaa.

Deebbii naaf Ergaa.

Answer we send?

To find the value or values of x for which the perimeters of the regular pentagon and the regular hexagon are equal, we need to set up an equation and solve for x.

Let's first calculate the perimeter of each polygon:

Perimeter of the regular pentagon = 5 * (x^2 + 17)
Perimeter of the regular hexagon = 6 * (x^2 + 2x)

Now we can set these two expressions equal to each other, as their perimeters are equal:

5 * (x^2 + 17) = 6 * (x^2 + 2x)

Simplify the equation:

5x^2 + 85 = 6x^2 + 12x

Rearrange the equation to get all terms on one side:

6x^2 - 5x^2 - 12x - 85 = 0

Combine like terms:

x^2 - 12x - 85 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = 1, b = -12, and c = -85. Substituting these values into the formula:

x = (-(-12) ± √((-12)^2 - 4*1*(-85))) / (2*1)

Simplifying:

x = (12 ± √(144 + 340)) / 2
x = (12 ± √484) / 2
x = (12 ± 22) / 2

We have two possible solutions:

1. x = (12 + 22) / 2 = 34 / 2 = 17
2. x = (12 - 22) / 2 = -10 / 2 = -5

Therefore, the values of x for which the perimeters of the regular pentagon and hexagon are equal are x = 17 and x = -5.