Question

(3x-1)(3x+1) = (x +2)(x-3)+14​

Answer 1

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Answer 2

To solve the equation (3x - 1)(3x + 1) = (x + 2)(x - 3) + 14, follow these steps:

1. Expand both sides of the equation: (3x - 1)(3x + 1) = (x + 2)(x - 3) + 14

   Expand the left side: 9x^2 - 1 = (x + 2)(x - 3) + 14

2. Expand the right side by multiplying (x + 2) and (x - 3) using the distributive property:

   9x^2 - 1 = (x * x + x * (-3) + 2 * x - 2 * 3) + 14

   Simplify the right side further: 9x^2 - 1 = (x^2 - 3x + 2x - 6) + 14

3. Combine like terms on the right side of the equation:

   9x^2 - 1 = (x^2 - x - 6) + 14

4. Now, simplify the equation by moving all terms to one side to set it equal to zero:

   9x^2 - 1 - x^2 + x + 6 - 14 = 0

5. Combine like terms:

   8x^2 + x - 9 = 0

6. The equation is now a quadratic equation in the form of ax^2 + bx + c = 0. You can solve this quadratic equation using the quadratic formula:

   x = (-b ± √(b^2 - 4ac)) / (2a)

   In this case, a = 8, b = 1, and c = -9. Plug these values into the formula:

   x = (-1 ± √(1^2 - 4 * 8 * (-9))) / (2 * 8)

   x = (-1 ± √(1 + 288)) / 16

   x = (-1 ± √289) / 16

7. Take the square root of 289:

   x = (-1 ± 17) / 16

8. Solve for the two possible values of x:

   x = (-1 + 17) / 16 = 16 / 16 = 1 x = (-1 - 17) / 16 = -18 / 16 = -9/8

So, the solutions to the equation are x = 1 and x = -9/8.