Question

(3x-3)^2-(x-15)^2=0​

Answer 1

$(3x - 3) {}^{2} - (x - 15) {}^{2} = 0 \\ (3x - 3 + x - 15)(3x - 3 - x + 15) = 0 \\ \left[ 4x - 18 = 0\\ 2x + 12 = 0 \right. \\ 4x = 18 \\ 2x = - 12 \\ x = 4.5 \\ x = - 6$

Ответ: -6, 4,5

Answer 2

To solve the equation $(3x - 3)^2 - (x - 15)^2 = 0$, you can follow these steps:

Step 1: Expand the squared terms on both sides of the equation. $(3x - 3)^2 - (x - 15)^2 = 0$ $9x^2 - 18x + 9 - (x^2 - 30x + 225) = 0$

Step 2: Distribute the negative sign on the right side. $9x^2 - 18x + 9 - x^2 + 30x - 225 = 0$

Step 3: Combine like terms. $8x^2 + 12x - 216 = 0$

Step 4: Divide the equation by any common factors to simplify it if possible. In this case, you can divide the equation by 4 to simplify it further. $2x^2 + 3x - 54 = 0$

Step 5: Factor the quadratic equation, or use the quadratic formula to find the solutions for $x$.

If we can factor the quadratic equation, we'll look for two numbers that multiply to give the product of the coefficient of $x^2$ (2) and the constant term (-54), and at the same time, add up to the coefficient of $x$ (3).

After trying different factor pairs, we find that $9$ and $-6$ satisfy these conditions: $2x^2 + 3x - 54 = (2x - 6)(x + 9) = 0$

Setting each factor equal to zero: $2x - 6 = 0 \quad \text{or} \quad x + 9 = 0$

Solving for $x$ in each equation: $2x = 6 \quad \text{or} \quad x = -9$

Dividing by 2: $x = 3 \quad \text{or} \quad x = -9$

So, the solutions to the equation $(3x - 3)^2 - (x - 15)^2 = 0$ are $x = 3$ and $x = -9$.